lib? 


IN   MEMORIAM 
FLOR1AN  CAJOR1 


(yr/fvt'aA*-   C*l^- 


r~- 


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in  2008  with  funding  from 

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DIFFERENTIAL  AND   INTEGRAL 
CALCULUS 


•?&&& 


DIFFERENTIAL  AND   INTEGRAL 
CALCULUS 

FOR 

TECHNICAL  SCHOOLS  AND    COLLEGES 


BY 

P.    A.    LAMBERT,   M.A. 

ASSISTANT   PROFESSOR  OF  MATHEMATICS,   LEHIGH  UNIVERSITY 


Neto  gork 
THE   MACMILLAN   COMPANY 

LONDON :  MACMILLAN  &  CO.,  Ltd, 

1898 

All  rights  reserved 


Copyright,  1898, 
By  THE  MACMILLAN  COMPANY. 


NorfoooH  $re88 

J.  S.  Cushing  &  Co.  —  Berwick  &  Smith 

Norwood  Mass.  U.S.A. 


PREFACE 

This  text-book  on  the  Differential  and  Integral  Calcu- 
lus is  intended  for  students  who  have  a  working  know- 
ledge of  Elementary  Geometry,  Algebra,  Trigonometry, 
and  Analytic  Geometry. 

The  object  of  the  text-book  is  threefold  : 

By  a  logical  presentation  of  principles  to  inspire  confi- 
dence in  the  methods  of  infinitesimal  analysis. 

By  numerous  problems  to  aid  in  acquiring  facility  in 
applying  these  methods. 

By  applications  to  problems  in  Physics  and  Engineer- 
ing, and  other  branches  of  Mathematics,  to  show  the  prac- 
tical value  of  the  Calculus. 

The  division  of  the  subject-matter  according  to  classes 
of  functions,  makes  it  possible  to  introduce  these  applica- 
tions from  the  start,  and  thereby  arouse  the  interest  of 
the  student. 

The  simultaneous  treatment  of  differentiation  and  inte- 
gration, and  the  use  of  trigonometric  substitution  to 
simplify  integration,  economize  the  time  and  effort  of  the 

student. 

P.   A.   LAMBERT. 

v 


K/«06070 


TABLE   OF   CONTENTS 


CHAPTER   I 
On  Functions 

ARTICLE  PAGE 

1.  Definition  of  a  Function 1 

2.  The  Indefinitely  Large  and  the  Indefinitely  Small     ...  2 

3.  Limits 3 

4.  Corresponding  Differences  of  Function  and  Variable        .        .  5 

5.  Classification  of  Functions         .        .  ■ 6 

CHAPTER   II 
The  Limit  of  the  Ratio  and  the  Limit  of  the  Sum 

6.  Direction  of  a  Curve 9 

7.  Velocity 12 

8.  Rate  of  Change 14 

9.  The  Limit  of  the  Sum 15 

10.  General  Theory  of  Limits 17 

11.  Continuity 18 

CHAPTER  III 
Differentiation  and  Integration  of  Algebraic  Functions 

12.  Differentiation 21 

13.  Integration 29 

14.  Definite  Integrals 34 

15.  Evaluation  of  the  Limit  of  the  Sum 36 

16.  Infinitesimals  and  Differentials 37 

CHAPTER  IV 
Applications  of  Algebraic  Differentiation  and  Integration 

17.  Tangents  and  Normals 43 

18.  Length  of  a  Plane  Curve 45 

vii 


Yin  CONTENTS 

ARTICLE  PAGE 

19.  Area  of  a  Plane  Surface 48 

20.  Area  of  a  Surface  of  Revolution 50 

21.  Volume  of  a  Solid  of  Revolution 52 

22.  Solids  generated  by  the  Motion  of  a  Plane  Figure     .        .        .53 

CHAPTER   V 
Successive  Algebraic  Differentiation  and  Integration 

23.  The  Second  Derivative 57 

24.  Maxima  and  Minima 63 

25.  Derivatives  of  Higher  Orders 74 

26.  Evaluation  of  the  Indeterminate  Form  - 75 

0 

CHAPTER   VI 

Partial  Differentiation  and  Integration  of 
Algebraic  Functions 

27.  Partial  Differentiation 78 

28.  Partial  Integration 82 

29.  Differentiation  of  Implicit  Functions 83 

30.  Successive  Partial  Differentiation  and  Integration     ...  86 

31.  Area  of  Any  Curved  Surface 89 

32.  Volume  of  Any  Solid 91 

33.  Total  Differentials     .........  92 

34.  Differentiation  of  Indirect  Functions 96 

35.  Envelopes 97 

CHAPTER   VII 
Circular  and  Inverse  Circular  Functions 

36.  Differentiation  of  Circular  Functions 


37.  Evaluation  of  the  Forms  oo  •  0,  co  —  co,  ~  . 

38.  Integration  of  Circular  Functions 

39.  Integration  by  Trigonometric  Substitution 

40.  Polar  Curves 

41.  Volume  of  a  Solid  by  Polar  Space  Coordinates 

42.  Differentiation  of  Inverse  Circular  Functions 

43.  Integration  by  Inverse  Circular  Functions 

44.  Radius  of  Curvature 


100 
105 
107 
111 
113 
117 
118 
120 
123 


CONTENTS 

ix 

CHAPTER  VIII 

Logarithmic  and  Exponential  Functions 

ARTICLE                                        .               ^  ^                                                                                                                            PAGE 

45. 

The  Limit  of  (  1  +  -  J   when  Limit  z  —  <x>         .         .         .         .128 

46. 

Differentiation  of  Logarithmic  Functions  . 

.     131 

47. 

Integration  by  Logarithmic  Functions 

.     133 

48. 

Integration  by  Partial  Fractions 

.     135 

40. 

Integration  by  Parts  .... 

130 

50. 

Integration  by  Rationalization  . 

142 

51. 

Evaluation  of  Forms  1*,  op0,  0° 

144 

52. 

Differentiation  of  Exponential  Functions 

145 

53. 

Integration  of  Exponential  Functions 

147 

54. 

The  Hyperbolic  Functions 

148 

55. 

The  Definite  Integral  f   "e-*'.  dx      . 

J—  00 

150 

56. 

Differentiation  of  a  Definite  Integral 

151 

57. 

Mean  Value 

152 

CHAPTER  IX 

Center  of  Mass  and  Moment  of  Inertia 

58. 

Center  of  Mass 154 

50. 

Center  of  Mass  of  Lines    . 

155 

60. 

Center  of  Mass  of  Surfaces 

157 

61. 

Center  of  Mass  of  Solids    . 

161 

62. 

Theorems  of  Pappus  .... 

162 

63. 

Moment  of  Inertia      .... 

165 

64. 

Moment  of  Inertia  of  Lines  and  Surfaces 

167 

65. 

Moment  of  Inertia  of  Solids 

170 

CHAPTER    X 
Expansions 

66.  Convergent  Power  Series 173 

67.  Taylor's  and  Maclaurin's  Series 175 

68.  Euler's  Formulas  for  Sine  and  Cosine 184 

69.  Differentiation  and  Integration  of  Power  Series        .         .         .186 

70.  Expansion  of  «i  =/  (x  +  h,  y  +  k) 192 


CONTENTS 


CHAPTER   XI 
Applications  of  Taylor's  Series 

ARTICLE  PAGE 

71.  Maxima  and  Minima  by  Expansion 194 

72.  Contact  of  Plane  Curves 198 

73.  Singular  Points  of  Plane  Curves 201 

CHAPTER   XII 
Ordinary  Differential  Equations  of  First  Order 

74.  Formation  of  Differential  Equations .        .  .        .         .     205 

75.  Solution  of  First  Order  Differential  Equations  of  First  Degree  .     209 

76.  Equations  of  First  Order  and  Higher  Degrees   .         .         .        .218 

77.  Ordinary  Equations  in  Three  Variables 222 

CHAPTER  XIII 
Ordinary  Differential  Equations  of  Higher  Order 

78.  Equations  of  Higher  Order  and  First  Degree     ....  225 

79.  Symbolic  Integration 230 

80.  Symbolic  Solution  of  Linear  Equations 234 

81.  Systems  of  Simultaneous  Differential  Equations        .        .        .  237 

CHAPTER  XIV 
Partial  Differential  Equations 

82.  Formation  of  Partial  Differential  Equations      ....     239 

83.  Partial  Differential  Equations  of  First  Order    .         .         .         .240 

84.  Linear  Equations  of  Higher  Order 244 


DIFFERENTIAL   AND    INTEGRAL 
CALCULUS* 

CHAPTER   I 

ON    FUNCTIONS 

Art.  1.  —  Definition  of  a  Function 

A  constant  is  a  quantity  which  retains  the  same  value 
throughout  a  discussion. 

A  variable  is  a  quantity  which  has  different  successive 
values  in  the  same  discussion. 

If  two  variables  are  so  related  that  to  every  value  of  the 
first  there  correspond  one  or  more  determinate  values  of  the 
second,  the  second  is  called  a  function  of  the  first. 

Denote  the  variables  by  x  and  y,  and  let  the  relation  between 
them  be  expressed  by  the  equation  y  =  mx  +  «,  where  m  and 
n  are  constants.  If  in  a  particular  discussion  m  =  3  and 
n  =  5,  the  equation  becomes  y  =  Sx-\-  5.  Arbitrary  values 
may  be  assigned  to  x,  and  the  corresponding  values  of  y  calcu- 
lated.    If  in  another  discussion  m  =  —  2  and  n  =  6,  the  equa- 

*  The  Differential  and  Integral  Calculus  was  invented  independently 
by  Newton  and  Leibnitz,  Newton  antedating  Leibnitz  by  several  years. 
Leibnitz  published  his  method  in  the  "Acta  Eruditorum"  of  Leipzig,  in 
1G84  ;  Newton  published  his  method  in  his  "  Natural  Philosophy,"  in  1687. 

B  1 


2  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

tion  becomes  y  =  —  2x  +  6.  Again,  arbitrary  values  may  be 
assigned  to  x,  and  the  corresponding  values  of  y  calculated. 

The  variable  x,  to  which  arbitrary  values  are  assigned,  is 
called  the  independent  variable.  The  variable  y,  whose  value 
depends  on  x,  is  called  the  dependent  variable  or  function. 

In  the  equation  y  =  Sx  -f  5,  if  x  increases,  y  increases  ;  if  x 
decreases,  y  decreases.  This  fact  is  expressed  by  calling  y  an 
increasing  function  of  x. 

In  the  equation  y  =  — - — ,  if  x  increases,  y  decreases :  if  x 
x  +  b 

decreases,  y  increases.  This  fact  is  expressed  by  calling  y  a 
decreasing  function  of  x. 

Art.  2.  —  The  Indefinitely  Large  and  Indefinitely 

Small 


The  term  of  the  geometric  progression  1,  2,  22,  23,  2\  25,  ... 
continually  increases,  and  becomes  larger  than  any  number 
that  can  be  assigned  when  the  progression  is  extended  suffi- 
ciently far. 

The  term  of  the  geometric  progression  1,  -,  — ,  — ,  — ,  — ,  •  •  • 

z  z    z    z    z 

continually  decreases  and  becomes  smaller  than  any  number 
that  can  be  assigned  when  the  progression  is  extended  suffi- 
ciently far. 

A  variable  quantity  whose  numerical  value  continually 
increases  and  becomes  larger  than  any  quantity  that  can  be 
assigned  is  said  to  become  indefinitely  large. 

A  variable  quantity  whose  numerical  value  continually 
decreases  and  becomes  smaller  than  any  quantity  that  can  be 
assigned,  is  said  to  become  indefinitely  small. 

3 

If  y  = -,  and  x  is   positive   and   becomes   indefinitely 

x-\-  5 

large,  the  corresponding  value  of  y  is  positive,  and  becomes 


ON  FUNCTIONS  3 

indefinitely  small.  If  x  is  negative,  and  starting  from  zero 
continually  approaches  —  5  in  such  a  manner  that  the  differ- 
ence between  the  value  assigned  to  x  and  —  5  becomes  indefi- 
nitely small,  the  corresponding  value  of  y  is  positive  and 
becomes  indefinitely  large.  If  x  continues  to  decrease  beyond 
—  5,y  becomes  negative  and  its  numerical  value  decreases, 
becoming  indefinitely  small  when  the  numerical  value  of  x 
becomes  indefinitely  large. 

All  quantities  which  lie  between  those  which  are  indefi- 
nitely small  and  those  which  are  indefinitely  large  are  called 
finite. 

If  e  denotes  an  indefinitely  small  quantity,  and  n  is  finite, 
the  product  n-e  must  also  be  indefinitely  small.  For,  if 
n  -  e  =  m,  and  m  is  finite,  e  =  — ,  the  ratio  of  two  finite  quanti- 
ties. Hence  c  would  be  a  quantity  whose  value  can  be  as- 
signed, which  is  contrary  to  the  hypothesis. 

Since  y  (x  +  5)  =  3  is  always  true  if  y  = -,  and  when  x 

becomes  indefinitely  large,  y  becomes  indefinitely  small,  it  fol- 
lows that  the  product  of  an  indefinitely  small  quantity  and  an 
indefinitely  large  quantity  may  be  finite. 

The  sum  of  a  finite  number  of  indefinitely  small  quantities 
is  indefinitely  small.  For  if  e  is  the  largest  of  the  n  indefi- 
nitely small  quantities  cj,  e2,  es,  e4,  •••,  en,  their  sum 

cannot  be  greater  than  n  •  c,  which  is  indefinitely  small  when 
n  is  finite. 

Art.  3.  —  Limits 

If  one  quantity  continually  approaches  a  second  quantity  in 
such  a  manner  that  the  difference  between  the  two  becomes 


4  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

indefinitely  small,  the  second  quantity  is  called  the  limit  of 
the  first. 

For  example,  if  y  =  5  +  -  when  x  becomes  indefinitely  large, 

3  x 

-  becomes  indefinitely  small,  and  by  the  definition  5  is  the 

x 

limit  of  y. 

The  limit  of.  a  quantity  which  becomes  indefinitely  small  is 
zero. 

The  limit  of  a  quantity  which  becomes  indefinitely  large  is 
called  infinity,  and  is  denoted  by  the  symbol  oo. 

This  conception  of  a  limit  is  used  in  elementary  geometry 
when  the  circumference  of  a  circle  is  proved  to  be  the  limit  of 
the  perimeter  of  the  inscribed  regular  polygon  when  the  num- 
ber of  sides  becomes  indefinitely  large ;  when  the  area  of  a 
circle  is  proved  to  be  the  limit  of  the  area  of  the  inscribed 
regular  polygon  when  the  number  of  sides  becomes  indefinitely 
large ;  when  the  volume  of  a  triangular  pyramid  is  proved  to 
be  the  limit  of  the  sum  of  the  volumes  of  inscribed  triangular 
prisms  of  equal  altitude  and  with  bases  parallel  to  the  base 
of  the  pyramid  when  the  number  of  prisms  becomes  indefi- 
nitely large. 

In  elementary  algebra,  the  sum  of  n  terms  of  the  geometric 
progression  a,  a  •  r,  a-r2,  a-r3,  a  «r4,  •••  is  proved  to  be 


a  —  a-i*         a         a-f 


1  —  r         1  —  r     1  —  r 

If  r  is  numerically  less  than  unity  and  a  is  finite,  a'r   be- 

1  —  r 
comes  indefinitely  small  when  n  becomes  indefinitely  large. 

Hence  when  n  becomes  indefinitely  large,  the  limit  of  sn  is, 

a 

l.-.f 


ON  FUNCTIONS  5 

If  r  is  positive,  the  successive  values  of  sn  as  n  increases  all 
lie  on  the  same  side  of  the  limit ;  if  r  is  negative,  the  succes- 
sive values  of  sn  oscillate  from  one  side  of  the  limit  to  the 
other.  For  example,  the  limit  of  the  sum  of  an  indefinitely 
large  number  of  terms  of  the  geometric  progression  1,  —  i,  J, 

1  1  1  1  1  1  1  !,„        ifl      J  TVlP       firQf 

¥>    TF>         "3~2">    6"¥>         T2  8>    "2  56"?         "3T2">     '         J    lb     3'         xne    niSt 

ten  successive  approximations  are  sx  =  1,  s2  =  i,  s3  =  |,  s4  =  |, 

Q    —  11         o    —  21        q    —  43        o    —    85  o    —  171        o     3  41         TTia 

approximation  sn  is   larger   than   the   limit   when  n  is  odd, 
smaller  than  the  limit  when  n  is  even. 

x2  —  1 

If  y  = -,  y  has  a  determinate  value  for  every  value  of 

OS  —  J. 

x,  except  for  x  =  1.     When  a  =  1,  ?/  takes  the  indeterminate 

form  -,  which  may  have  any  value  whatever.     The  true  value 

of  y  when  x  =  1  is  defined  as  the  limit  of  the  values  of  y  cor- 
responding to  values  of  x  whose  limit  is  1.     For  example, 

when  0  =  1.1,  1.01,  1.001,  1.0001,  1.00001,  1.000001,  •••, 

y  =  2.1,  2.01,  2.001,  2.0001,  2.00001,  2.000001,  .... 

Hence  2  is  the  true  value  of  y  when  x  =  1,  and  for  all  values 

rjfi  i 

of  x,  including  x  =  1,  y  = =  x  +  1. 


Art.  4.  —  Corresponding  Differences  of  Function 
and  Variable 

If  y  =  a2,  when  g  =  -  4,   -  3,  -  2,  -  1,    0,    1,    2,    3,     4, 
2/=   16,       9,       4,       1,    0,    1,    4,    9,  16. 

Starting  from   x  =  0,  a  difference  of   +1  in  the  value  of  x 
causes  a  difference  of   -f  1  in  the  value  of  y ;  starting  from 


6  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

x  =  -f- 1,  a  difference  of  +  1  in  the  value  of  x  causes  a  differ- 
ence of  -h  3  in  the  value  of  y ;  starting  from  x  =  —  3,  a  dif- 
ference of  -f  1  in  the  value  of  x  causes  a  difference  of  —  5  in 
the  value  of  y.  Observe  that  the  same  change  in  the  value  of 
the  variable  in  different  parts  of  the  function  causes  different 
changes  in  the  value  of  the  function. 

In  general,  if  y  =  x2,  and  starting  from  any  value  of  x  the 
corresponding  differences  of  y  and  x  are  denoted  by  Ay  and 
Ax,  so  that  x  +  Ax  and  y  -\-  Ay  must  satisfy  the  equation 
y  =  x2,  by  subtracting  the  equations  y  +  Ay  =  (x  +  Ax)2  and 
y  =  x2,  there  results  Ay  =  2  x  •  Ax  4-  (Ax)2.  The  difference  in 
the  value  of  y  corresponding  to  a  difference  of  Ax  in  the  value 
of  x  is  seen  to  depend  on  x  and  on  Ax. 


Art.  5.  —  Classification  of  Functions 

A  function  is  called  algebraic  if  the  relation  between  func- 
tion and  variable  can  be  expressed  by  means  of  a  finite  number 
of  the  fundamental  operations  of  algebra,  addition,  subtrac- 
tion, multiplication,  division,  and  involution  and  evolution 
with  constant  indices. 

For  example,  y  =  x*  —  lx  +  1  explicitly  defines  y  as  an 
integral,  rational,  one-valued,  algebraic  function  of  x.  The 
equation  x2  -f  y2  —  9  implicitly  defines  y  as  a  two-valued  alge- 
braic function  of  x.    The  relation  y  =  — ^--  defines  y  as  a 

1  +  ff* 
fractional,  irrational,  one-valued  function  of  x. 

All  functions  not  algebraic  are  called  transcendental.  The 
expression  of  transcendental  functions  by  means  of  the  funda- 
mental operations  of  algebra  is  possible  only  in  the  form  of 
the  sum  of  an  indefinitely  large  number  of  terms,  or  in  the 
form  of  the  product  of  an  indefinitely  large  number  of  factors. 


ON  FUNCTIONS  7 

The  elementary  transcendental  functions  are  : 
The  exponential  function  y  =  ax  and  its  inverse  the  loga- 
rithmic function  x  =  logay. 

The  trigonometric  or  circular  functions  y=  sin  a;,  y  =  tan  x, 
y  =  sec  x,  together  with  their  complementary  functions ;  and 
the  inverse  functions  x  =  sm~Ay,  x  =  tdirrly,  x=sec~1y. 

In  general,  the  fact  that  y  is  an  explicit  function  of  x,  with- 
out specifying  the  nature  of  the  function,  is  denoted  by  writing 
y  =f(x),  or  y  =  F(x),  or  y  =  <£  (x) ;  the  fact  that  y  is  an  im- 
plicit function  of  x  is  denoted  by  writing  f(x,  y)  =  0,  or 
F(x,y)  =  0,  or  <f>(x,  y)  =  0. 

PROBLEMS 

1.  Determine  the  values  of  the  function  y  =  3x2—  5  cor- 
responding to  x  =  0,  1,  2,  3,  4,  5. 

x2  —  A 

2.  Find  the  true  value  of  y  = when  x  =  2. 

x  —  2 

a?  —  1 

3.  Find  the  true  value  of  y  = when  x  =  l. 

x  —  1 

4.  Starting  from  x  =  3,  calculate  the  difference  in  y  cor- 
responding to  a  difference  of  2  in  the  value  of  x  if 

y  =  x2  -5x  +  12. 

5.  Starting  from  x  =  2,  calculate  the  difference  in  y  corre- 
sponding to  a  difference  of  —2  in  the  value  of  x  \fy  =  1  x  —  3x*. 

6.  Find  the  limit  of  the  sum  of  an  indefinitely  large  num- 
ber of  terms  of  the  series  1  +  \  +  -J-  +  <fa  +  -fa  H . 

7.  Find  the  limit  of  the  sum  of  an  indefinitely  large  num- 
ber of  terms  of  the  series  ft  +  T^  +  T^  +  TTFihnF  + -•••• 


8  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

8.  Show  that  any  difference  in  the  abscissa  of  the  straight 
line  whose  equation  is  y  =  mx  +  n  causes  a  difference  m  times 
as  large  in  the  ordinate. 

9.  Show  that  the  ordinate  of  the  straight  line  whose  equa- 
tion is  y  —  yo  =  m(x  —  x0)  changes  m  times  as  fast  as  the 
abscissa. 

10.  If  y  =  Sx2  —  7 x,  determine  the  change  Ay  in  the  value 
of  y  corresponding  to  a  change  of  Ax  in  the  value  of  x. 

11.  Compare  the  values  of  Ay  corresponding  to  A#=l, 
starting  from  x  =  0,  1,  2,  3,  if  y  =  x2  —  3#  +  10. 

12.  If  y  =  ax2  +  bx  +  c,  where  a,  b,  and  c  are  constants  and 
when  x  =  x0,  y  —  y0,  determine  the  change  Ay  in  the  value  of  y 
corresponding  to  a  change  of  Ax  in  the  value  of  x,  starting 
from  x  as  x0. 


CHAPTER   II 


THE  LIMIT  OF  THE  KATIO  AND  THE  LIMIT  OP  THE  SUM 


Art.  6.  —  Direction  of  a  Curve 


Let  (x0,y0)  be  any  point  of  the  curve  whose  equation  is 
y  =  x2.  Let  (x0  +  Ax,  y0  -f-  Ay)  be  any  other  point  of  the  curve, 
Ax  and  Ay  representing  corresponding  differences  in  abscissa 

and  ordinate.     The  ratio  — ^  is  the  slope  of  the  secant  line 

Ax  r 

through  (x0,  y0)  and  (x0  -f-  Ax,  y0  +  Ay),  that  is,  the  tangent  of 
the  angle  of  inclination  of  the 
secant  to  the  X-axis.    The  equa- 
tion of  the  secant  is 


V 


Ax 


x0). 


Fio.l. 


This  is  true  whatever  may  be 
the  magnitude  of  the  corre- 
sponding differences  Ax  and 
Ay. 

Now  the  tangent  to  a  curve 
is  defined  as  the  limiting  posi- 
tion   of   the   secant   whose    two   points   of   intersection  are 
made   to   continually   approach    each    other.      If    the    point 
(x0  +  Ax,  yQ  -f-  Ay)  continually  approaches  the  point  (a?0,  y0),  the 
limit  of  Ax  is  zero,  and  the  corresponding  limit  of  the  ratio  — J- 


10  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

is  the  slope  of  the  tangent  to  the  curve  at  (x0,  y0).  This  limit 
of  the  ratio  is  to  be  determined. 

By  hypothesis,  y0+Ay  =  (xQ  +  Ax)2,  yQ  =  x02 ;  whence 

Aff  _  (xo  +  Aa02  —  xo2. 
Ax  Ax 

The  limit  of  this  ratio,  when  the  limit  of  Ax  is  zero,  cannot 
be  found  by  placing  Ax  =  0  in  this  value  of  the  ratio.  For 
this  makes  the  ratio  take  the  indeterminate  form  -  as  it  ought 

to,  for  the  two  points  are  made  coincident,  and  through  one 
point  an  infinite  number  of  straight  lines  may  be  drawn.  By 
performing  the  operations  indicated  in  the  numerator  of  the 

value  of  the  ratio  —  and  then  dividing  out  the  factor  Ax  com- 
Ax  b 

mon  to  numerator  and  denominator,  there  results 

-^-  =  2xq-\-Axq. 
Ax 

If  now  Ax  becomes  indefinitely  small,  that  is,  if  the  limit  of 

Ax  is  zero,  the  limit  of  the  ratio  — ^  is  2  xn. 
'  Ax  ° 

Denoting  by  a  the  angle  of  inclination  to  the  X-axis  of  the 
tangent  to  y  =  x2  at  (x0,y0),  tan«  =  2x0  and  the  equation  of 
the  tangent  is  y  —  y0  =  2x0(x  -  x0).  At  the  point  (2,4), 
tan  a  =  4  and  «  =  75°  58'.  The  tangent  makes  an  angle  of  45° 
with  the  X-axis  if  2x0  =  tan  45°  =  1.  Solving  the  equations 
2  x0  =  l  and  yQ=x02,  the  point  of  tangency  is  found  to  be 
»o  =  |>  #o  =  i- 

If  the  curve  whose  equation  is  y  =  x2  is  generated  by  the 
continuous  motion  of  a  point,  when  the  generating  point  passes 
the  point  (x0,  y0)  of  the  curve  it  tends  at  that  instant  to  move 
along  the   tangent  y-y0=  2x0(x  —  x0)   at  the   point   (xQ,y0). 


THE  LIMIT  OF  THE  RATIO  11 

Hence  the  direction  of  the  tangent  to  the  curve  at  any  point 
is  called  the  direction  of  the  curve  at  that  point. 

If  a  point  moves  along  the  straight  line  y  —  y0  =  2 x0(x  —  x0), 
the  ordinate  changes  2x0  times  as  fast  as  the  abscissa.  When 
2#0is  positive,  the  ordinate  is  an  increasing  function  of  the 
abscissa;  when  2x0  is  negative,  the  ordinate  is  a  decreasing 
function  of  the  abscissa ;  when  2  a?0  =  0,  the  line  is  parallel 
to  the  X-axis  and  a  change  in  the  abscissa  causes  no  change 
in  the  ordinate. 

Hence,  when  the  point  generating  the  curve  y  =  x2  passes 
the  point  (x0,  y0),  the  ordinate  of  the  curve  is  at  that  instant 
changing  value  2x0  times  as  fast  as  the  abscissa  changes.  At 
the  point  (J,  \)  ordinate  and  abscissa  are  changing  value  at 
the  same  rate  ;  at  the  point  (2,  4)  the  ordinate  is  increasing  4 
times  as  fast  as  the  abscissa  increases;  at  the  point  (—2,4) 
the  ordinate  decreases  4  times  as  fast  as  the  abscissa  increases. 

By  precisely  the  same  analysis  it  is  proved  that  the  slope  of 
the  tangent  to  the  curve  whose  equation  is  y  =/(»)  at  any  point 

(x,  y)  of  the  curve  is  tan  a  =  limit  —  =  limit  — — — — -* — "-S 
v  '  9J  Ax  Ax 

when  the  limit  of  Ax  is  zero,  and  that  the  limit  of  this  ratio 
measures  the  rate  of  change  of  ordinate  and  abscissa  at  (x,  y). 
It  is  essential  to  remember  that  in  the  calculation  of  this 
limit  Ax  must  start  from  some  finite  value  and  then  be  made 
to  approach  the  limit  zero.* 

PROBLEMS 

1.  Find  the  slope  of  the  tangent  to  y  =  x3  —  Sx  at  x  =  5. 

2.  Find  the  direction  in  which  the  point  generating  the 
graph  of  y  =  Zx2  —  x  tends  to  move,  when  x  =  1. 

*The  tangent  problem  prepared  the  way  for  the  invention  of  the 
Differential  Calculus,  in  the  seventeenth  century. 


12  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

3.  Find  the   rate  of  change  of  ordinate  and  abscissa  of 
y  =  Sx2  —  x  at  x  =  l. 

4.  Find   at   what  point  of  the   curve  whose   equation  is 
y  =  4  x2  the  tangent  makes  with  the  X-axis  an  angle  of  45°. 

5.  Find  the  equation  of  the   tangent  to  y  =  2x2  —  5x  at 
x=3. 

6.  Find   where   the   ordinate  of  y  =  Sx  —  4#2  decreases  5 
times  as  fast  as  x  increases. 


Art.  7.  —  Velocity 

Suppose  a  locomotive  to  start  at  station  A,  to  pass  station  B 
distant  s0  miles  from  A  after  t0  hours,  and  station  C  distant 
s  miles  from  A  after  t  hours.     The  average  velocity  per  hour 


--So *i  C 


.§ j 

to  t\ 


Fig.  2. 


from  B  to  C,  that  is,  the  uniform  number  of  miles  per  hour 
the  locomotive  must  run  from  B  to  C  to  cover  the  distance 

s  —  s0  miles  in  t  —  tQ  hours,  is  s~  °.     Calling  the  difference  of 

t  —  £0 

distance  As  and  the  difference  of  time  A£,  the  average  velocity 

is  — .      The  equal  ratios    —  = 9  determine  the  average 

A*  *  At      t-t0  5 

velocity  of  the  locomotive  during  the  interval  of  time 
At  =  t  — 10,  whatever  may  be  the  magnitude  of  this  interval 
of  time. 

Now  if  station  O  is  taken  nearer  and  nearer  station  B,  the 
interval  of  time  At  becomes  indefinitely  small  and  has  zero 


THE  LIMIT  OF  THE  BATIO  13 

for   limit.     The   average   velocity   from  B  to   C  continually 
approaches  the  actual  velocity  at  B,  since  the  interval  of  time 

during  which  a  change  of  velocity  might  take  place  continu- 

As 

ally  decreases.     Hence  the  limit  of  the  ratio  —  when  the 

...  At 

limit  of  At  is  zero   is   the   actual  velocity  of  the  locomotive 

at  J3. 

This  analysis  shows  that  if  the  relation  between  distance  s 
and  time  t  of  the  motion  of  a  body  is  expressed  by  the  equa- 
tion s  =f(t),  and  the  velocity  at  any  time  t  is  denoted  by  v, 

v  =  limit  r  \  "*" — )  ~J\  )  when  the  limit  of  At  is  zero. 
At 
For  example,  in  the  case  of  a  freely  falling  body,  starting 

from  rest,  s  — 16.08 t2,  where  s  is  distance  measured  in  feet, 

and  t  is  time  measured  in  seconds.     Here 

v  =  limit  16.08(*  + A*)'- 16.08*' 

At 

=  limit  16.08  (2 1  +  At)  =  32.16 1, 

when  the  limit  of  A*  is  zero.     Hence  the  velocity  at  the  end  of 
the  third  second  is  96.48  feet  per  second. 

Velocity  is  seen  to  be  the  rate  of  change  of  distance  per 
unit  of  time. 

PROBLEMS 

1.  If  s  =  %gt2,  where  g  is  a  constant,  determine  the  velocity 
at  time  t. 

2.  If  s  =  10 £  + 16.08 12,  calculate  the  velocity  at  time  t. 

3.  If  s  =  ut  —  ^gt2,  where  u  and  g  are  constants,  determine 
the  velocity  at  time  t. 

4.  If  s  =  ut  +  \gt2,  where  u  and  g  are  constants,  determine 
the  velocity  at  time  t. 


14  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Art.  8.  —  Rate  of  Change 

In  the  function  y  =  x2  —  5  x  let  x0,  y0  and  x0  -f  Ax,  y0  +  Ay 
be  two  sets  of  corresponding  values  of  variable  and  function. 
That  is,  starting  from  x0,  y0  to  a  change  of  Ax  in  the  value  of 
the  variable  there  corresponds  a  change  of  Ay  in  the  value 
of  the  function.     The  ratio  of  the  corresponding  changes  of 

function  and  variable  — ^  determines  the  average  rate  of  change 

Ax 

of  the  function  throughout  the  interval  Ax;  that  is,  the  uni- 
form change  of  the  function  for  change  of  the  variable  by 
unity  which  in  the  interval  Ax  causes  a  change  of  Ay  in  the 
value  of  the  function.  This  is  true  for  all  values  of  the 
interval  Ax. 

Now,  if  Ax  becomes  smaller  and  smaller,  the  ratio  — '-  con- 
'  '  Ax 

tinually  approaches  the  actual  rate  of  change  of  the  function 
at  x0,  since  the  interval  Ax  during  which  the  rate  of  change 
might  vary  continually  decreases.  Hence  the  actual  rate  of 
change  of  the  function  y  =  x2  —  5x  at  x0,  yQ  is 

limit  ^L  =  limit  Oo  +  *XY  ~  5 (xo  +  Aft)  -  (x,2 -5x0) 

Ax  Ax 

=  limit  (2x0  —  5  +  Ax)  =  2xQ  —  5, 

when  the  limit  of  Ax  is  zero. 

The  function  y  =  x2  —  5x  increases  3  times  as  fast  as  x 
increases  when  2  x0  —  5  =  3,  that  is,  when  xQ  =  4 ;  the  function 
decreases  5  times  as  fast  as  x  increases  when  2a*0  — 5  =  — 5, 
that  is,  when  x0  =  0 ;  the  function  is  stationary,  that  is,  it  is 
neither  increasing  nor  decreasing,  when  2  x0  —  5  =  0. 
.  This  analysis   shows  that  for  any  value  of  x  the  limit  of 

•'  ^x  *" X~~J{X)  when  the  limit  of  A#  is  zero,  measures  the 

Aa; 


THE  LIMIT   OF  THE  SUM 


15 


rate  of  change  of  f(x)  for  that  value  of  x.     If  this  limit  is 

positive,  f(x)  is  an  increasing  function  of  #;  if  this  limit  is 

negative,  f(x)    is  a  decreasing  function  of   x\    if   this  limit 

is  zero,  f(x)  is  stationary. 

The  calculation  of  the  limit  of  the  ratio  /(«+Aa?)-/(a?) 

Ax 
when  limit  Ax  =  0  for  all  functions  f(x)  is  the  fundamental 

problem  of  the  Differential  Calculus. 

PROBLEMS 

1.  Calculate  the  rate  of  change  of  Ax  +  7. 

2.  Find  the  rate  of  change  of  y  =  Xs  +  Sx. 

3.  Calculate  the  rate  of  change  of  Sx  —  x2  at  a?  =  1. 

4.  Find  where  the  function  x2  —  2x  increases  twice  as  fast 
as  x  increases. 

5.  Find  where  the  function  x2  —  2x  decreases  twice  as  fast 
as  x  increases. 

6.  Find  where  the  function  x2  —  2x  is  stationary. 


Art.  9.  —  The  Limit  of  the  Sum 

Let  y=f(x)  be  the  equation  of  the  given  curve.  Denote 
by  A  the  area  of  the  surface 
bounded  by  the  curve,  the 
X-axis,  and  the  lines  x  =  a, 
x  —  b.  Divide  the  portion  of 
the  X-axis  from  x  =  a  to  x  =  b 
into  any  number,  say  5,  of 
equal  parts,  and  call  each  part 
Ax.  Constructing  rectangles  on 
each  Ax,  as  indicated  in  the 
figure,  and  denoting  by  A'  the 


16 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


sum  of  the  areas  of  the  rectangles, 

A'  =/(a)  •  Ax  4-/(0  +  Ax)  •  Ax  +/(a  +  2  Ax) .  Ax 
+/(&  -  2  Ax)  •  Ax  +f(b  —  Ax)  •  Ax, 


which  may  be  written  A'  =  2/(x)  •  Ax.     Now,  as  the  number 

x  =  a 

of  equal  parts  into  which  b  —  a  is  divided  is  indefinitely  in- 
creased,   Ax   becomes   indefinitely  small,  and  A'  continually 

3=6 
approaches  A.     Hence  A  —  limit  of  %/(%)  •  Ax  when  the  limit 

x  =  a 

of  Ax  is  zero.     The  calculation  of  this  limit  of  the  sum  is  one 
of  the  fundamental  problems  of  the  Integral  Calculus.* 

In  some  simple  problems 
the  limit  of  the  sum  may 
be  calculated  by  means  of 
the  formula  for  the  sum 
of  n  terms  of  an  arithmetic 
progression 

a+(a+ d)  -f  (a  +  2d)  +••• 

+  (l-2d)  +  (l-d)  +  l, 

namely,  $=(a  +  t)^- 

Fig.  4. 

For  example,  let  it  be 
required  to  calculate  the  area  of  the  surface  bounded  by  the 
straight  lines  y  =  x  +  2,  x  =  1,  x  =  5,  and  the  X-axis.  Divid- 
ing the  distance  from  x  =  1  to  x  =  5  into  n  equal  parts  and 

calling  each  part  Ax,  Ax  =  -,  and 


*  Historically  the  calculation  of  the  areas  of  surfaces  bounded  by 
curved  lines*  led  to  the  invention  of  the  Integral  Calculus. 


THE  LIMIT  OF  THE  SUM  17 


2  =  4 

A  =  limit  2/(#)  Ax 
1=1 

-iw{3+(.+g+(.+g+(»+2)+-~ 

=  limit  J6  +  (n-l) -!?.-  =  limit  2^10 --^  =  20 
(  n )  2  n  \         nj 

when  n  is  indefinitely  increased.  This  agrees  with  the  result 
obtained  by  elementary  geometry. 

Art.  10.  —  General  Theory  of  Limits 

Let  u  denote  a  function  of  x  whose  limit  is  U  when  the 
limit  of  x  is  a.  This  relation  may  be  denoted  by  the  equation 
u(x=ak8)  =  U±  c,  where  e  must  become  indefinitely  small 
when  8  becomes  indefinitely  small. 

Limit  of  the  sum.  —  Suppose  that  when  the  limit  of  x  is 
a,  limit  ux  =  U\,  limit  u2  =  U2)  limit  n3  =  TJZ.  The  hypothesis 
is  equivalent  to 

Ul(x  =  a±&)  =  R  ±  Clj     ^2(a:  =  a±S)  =  C/g  ±  62,     W3  (x  =  a  ±  j)  =  U3  ±  €3, 

whence  (ux  +  w2  —  w3)(x=a  ±  n  =  £7i  -f  {72  —  C/"3  ±  «!  ±  e2  T  c3-  Since 
when  8  becomes  indefinitely  small,  e1?  e2,  c3  each  become  indefi- 
nitely small,  ±  cj  ±  c2  =F  €3  for  all  combinations  of  signs  also 
becomes  indefinitely  small.  Hence  when  the  limit  of  x  is  a, 
limit  (ux  +  u2  —  w3)  =  Ui  +  Ui—Us  =  limit  ux  +  limit  u2  —  limit  w3. 
That  is,  the  limit  of  the  algebraic  sum  of  a  finite  number  of 
quantities  is  the  like  algebraic  sum  of  their  limits. 

Limit  of  the  product.  —  If,  when  the  limit  of  x  is  a,  limit 
^  =  ^1  and  limit  u2  =  U2,  «**<»«•**)  =  Ut  ±  €%  and  u2(x=a±s) 
=  U2±  «2-     By  multiplication, 

(iii  •  w^){,„#iW  =  Ui  •  U2  ±  U2  •  e,  ±  Ux  •  c2. 
c 


18  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Hence  if  Ui  and  U2  are  finite,  when  the  limit  of  x  is  a,  limit 
(id  •  u2)  =  Ui-  U2  =  limit  ux  •  limit  u2.  That  is,  the  limit  of  the 
product  is  the  product  of  the  limits. 

Limit  of  the  quotient. — If,  when  the  limit  of  x  is  a,  limit 
uL  =  Ux  and  limit  u2  =  U2,  u1^tmtl±Si^U1±€h  u2{x^aiz^  =  U2±e2. 
By  division, 

VV(x  =  a±S)  C^2±«2  ^2         U2  ±   C2         U2 

=  U1,±U2j_^TU11t2 

U2         U2(U2±c2) 

Hence    if   R  and    U2  are  finite,  when  the  limit  of  x  is  a, 

limit -1  =  -^^hlmt^1.     That  is,  the  limit  of  the  quotient  is 
u2      U2     limit  u2 

the  quotient  of  the  limits.* 


Art.  11.  —  Continuity 

The  function  y=f(x)  is  said  to  be  continuous  at  x  —  x0  if 
the  limit  of  the  difference  f(x0  ±  Ax)  —f(x0)  is  zero  when  the 
limit  of  Ax  is  zero.  This  definition  may  also  be  written 
[/(aJb  +  Aa;)  —  f(x0)^\Ax==:kS  =  ±  e,  where  c  must  become  indefi- 
nitely small  when  8  becomes  indefinitely  small.  The  function 
is  said  to  be  discontinuous  at  x  =  x0  if  e  does  not  become 
indefinitely  small  when  8  becomes  indefinitely  small. 

For  example,  if  the  curve  in  the  figure  is  the  graph  of 
y  =f(x),f(x)  is  continuous  at  all  points  except  at  x0  =  l  and 

*  Jordan,  in  his  Cours  d' analyse,  Paris,  1893,  perhaps  the  most  com- 
plete treatise  on  the  Calculus  ever  written,  says  :  "Arithmetic  and  Algebra 
employ  four  fundamental  operations,  addition,  subtraction,  multiplica- 
tion, and  division.  A  fifth  can  be  conceived  of,  consisting  in  replacing 
a  variable  quantity  by  its  limit.  It  is  the  introduction  of  this  new  opera- 
tion that  characterizes  the  Calculus." 


THE  LIMIT  OF  THE  SUM 


19 


/ 


330  A 

-4c 


;n,2) 


x0  =  —  3.  Starting  from  (1,  —  2),  f(x)  is  continuous  for  in- 
creasing values  of  x ;  starting  from  (1,  2),  f(x)  is  continuous 
for   decreasing   values    of    x.  Y 

Starting    from    (1,  2),    limit 
[/(l  +  Ao;)-/(l)]=4  when 
limit   Ax  =  0.      Starting  from 
(1,-2), 
limit[/(l-Aa)-/(l)]  =  -4 

when  limit  Ax  =  0.     Starting 
from  (—3,  oo), 

limit[/(-3  +  Aa;)-/(-3)] 

=  —  CO  Fig.  5. 

when  limit  Ax  =  0.  Starting  from  (—3,  -co),  limit 
[/(-  3  -  Ax)  -/(-  3)]  =  +  oo  when  limit  Ax  =  0. 

Starting  from  (1,  2), 

limit  /(i-Aa)-./Xl)  _  tan  60o     when  limit  Ax  =  0 

—  Ace 


00 


when  limit  Ax  =  0. 


limit /fl  +  ^WW 

Ax 

Starting  from  (1,  —  2), 

limit  /(1  +  Aa;)-/(1)  =  tftn  33()o  when  limit  Aa.  =  Q 

Aa? 


limit 


/(l  -  A*)  -/(l) 


Aa? 


CO 


when  limit  Ax  =  0. 


Hence  it  is  evident  that,  at  points  of  discontinuity  of  f(x), 

the  limit  of  the  ratio  ^  +  Aa?)  ~^°)  when  limit  Aa  =  0 

A# 

is  not  independent  of  the  algebraic  sign  of  Ax.  At  points  of 
continuity,  the  limit  of  this  ratio  generally  is  independent 
of  the  sign  of  Ax. 


20  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Example.  —  Show  that  log  x  is  a  continuous  function  of  x. 
Here   limit  [log  (x  ±  Ax)  —  log  x~]  =  limit  logf  1  ±  —  1  =  0, 
when  limit  Ax  =  0,  except  for  x  =  0. 

PROBLEMS 

Show  that  the  following  are  continuous  functions  of  x  : 

1.    3x?  —  5a +  2.  2.    sinsc.  3.    ex. 

Find  the  points  of  discontinuity  of 

4.    5^±1.  5.    tana.# 

5x-T 


CHAPTER   III 

DIFFEKENTIATION  AND   INTEGRATION   OF  ALGEBRAIC 
FUNCTIONS 

Art.  12.  —  Differentiation 

The  function  of  x  which  is  the  limit  of  the  ratio 

f(x  +  Ax)-f(x) 

Ax 

when  limit  Ax  =  0,  is  called  the  first  derivative  of  f(x)  with 

respect  to  x,  and  is  denoted  by  the  symbol  —  /(#)>  ov  /'(«). 

If  f(x)  is  denoted  by  y,  the  first  derivative  is  denoted  by  -^« 

The  operation  of  forming  the  first  derivative,  denoted  by  the 

symbol  — ,  is  called  differentiation.      General   rules   for  the 
dx 

differentiation  of  algebraic  functions  are  to  be  established. 

I.  Let  u  represent  any  continuous  function  of  x.  Represent 
by  Au  the  change  in  the  value  of  u  corresponding  to  a  change 

flti  /A  7/ 

of  Ax  in  the  value  of  x.     By  definition,  —  =  limit  —  when 

dx  Ax 

limit  Ax  =  0. 

*The  notation  ^  was  invented  by  Leibnitz  (1646-1716).  Newton 
(1642-1727)  denotes  —  by  s,  a  notation  still  used  in  mechanics.  La- 
grange (1736-1813)  denotes  the  first  derivative  of  f(x)  by/(x),  Cauchy 
(1789-1857)  by  Df(x). 

21 


22  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

II.    Let  u  =  c,  where  c  is  a  constant,  that  is,  a  change  in  the 
value  of  x  does  not  cause  a  change  in  the  value  of  c.     Then 

■ —  =  —  =  limit  —  =  0    when    limit  Ax  =  0.      Conversely,    if 
dx      dx  Ax 

—  =  0,  that  is,  identically  zero,  u  is  independent  of  x.     For 

O.X 

—  =  0   means   that  a  change  in  x  causes   no   change   in  u. 
dx 

Hence  u  is  independent  of  x. 

o 

dx     dx  Ax 


III.    Let  u  =  x.      Then   —  =  —  =  limit  —  =  1    when  limit 


Ax  =  0. 

IV.  Let  /(x)  =  C'U,  where  c  is  a  constant  and  u  represents 
a  continuous  function  of  x.     Forming  the  first  derivative, 

— f(x)  =  limit  — i — : '- =  limit  c =  c 

da;  Ax  Ax  dx 

when  limit  Ax  =  0.  Hence  the  first  derivative  of  a  function 
multiplied  by  a  constant  is  the  constant  times  the  first  deriva- 
tive of  the  function. 

V.  Let  /(x)  =  u  +  v  —  w,  where  u,  v,  iv  represent  continuous 
functions  of  x.  Denoting  by  Au,  Av,  and  Aw  the  changes  in 
the  values  of  u,  v,  and  w,  corresponding  to  a  change  Ax  in  the 
value  of  x, 

A  f(x\  _  limit u  +  Au  +  v  +  A^  ~  w  ~  Aw  ~  ^  +  v  ~  ?<?) 
daTW  Ax 

=  limit  S«  +  limit  **  -  limit  *5  =  **  +  ft  -  ^ 
Ax  Ax  Ax      dx     dx      dx 

when  limit  Ax  =  0.  That  is,  the  first  derivative  of  the  alge- 
braic sum  of  a  finite  number  of  functions  is  the  like  algebraic 
sum  of  the  first  derivatives  of  the  functions. 

If  two  functions  f(x)  and  <£  (x)  have  the  same  first  deriva- 


DIFFERENTIATION  AND  INTEGRATION  23 

tive,  their  difference  f(x)  —  cj>  (x)  is  a  constant.    By  hypothesis, 

hence  f(x)  —  c/>  (x)  =  c. 

VI.  Let  /(a;)  ss  m  •  v,  where  w  and  v  represent  continuous 
functions  of  x. 

—  f  (a;)  =  —  (w  •  V)  aa  limit  * — — '    v       ^ 

daTw      dxK       J  Ax 

=  limit  (it  -f  Ait) 1-  limit  v —  u \- v  — 

Aa;  Aa;  dx  dx 

when  limit  Aa;  =  0.     Hence  the  first  derivative  of  the  product 
of  two  functions  is  the  first  function  times  the  first  derivative 
of  the  second  function  plus  the  second  function  times  the  first 
derivative  of  the  first  function. 
In  like  manner  it  is  proved  that 

d  /  s  dw  .  dv  .  du 

{U'V'W)  —U'V \-U'W \-V'W 

dx  dx  dx  dx 

VII.  Let  f(x)  =  - ,  where  u  and  v  are  continuous  functions 

01  X. 

u  +  An     u 

g^/(^)^-4■ffl-^M^i^  +  *,    ■« 


dx  dx  \vj  Aa; 

Aw  Av  du  dv 

V U V' u 

..    ,,      Aa;           Aa;           dx           dx 
=  limit — 2— as 

ir  +  v  •  Av  vl 

when  limit  Aa;  =  0.  That  is,  the  first  derivative  of  the  quo- 
tient of  two  functions  is  the  divisor  times  the  first  derivative 
of  the  dividend  minus  the  dividend  times  the  first  derivative 
of  the  divisor,  divided  by  the  square  of  the  divisor. 


24  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

_      dv 

If  f(x)  =  -,  where  c  is  a  constant,  — [  -  )  = 5 Hence 

J  v  J      v  dxyvj  v2 

the  first  derivative  of  a  fraction  whose  numerator  is  constant 
and  whose  denominator  is  a  function  of  x,  is  the  numerator 
with  its  sign  changed  times  the  first  derivative  of  the  denomi- 
nator divided  by  the  square  of  the  denominator. 

VIII.    Let  f(x)  =  u11,  where  u  is  a  continuous  function  of  x, 
and  n  is  any  finite,  positive  integer. 


£.  f(x\  «-*«•«  limit 

dxJ K  }      dx 

(u  -f  Aw)M  —  un 

Ax 

=  limit  -j  n  « un' 

[f 

4- »  •  u  •  (A%)"-2  +  (A?*)"-1 

„  i   du 
dx 

Aw 
Ax 


when  limit  Asc  =  0.  Hence  the  first  derivative  of  a  function 
affected  by  a  finite,  positive,  integral  exponent  is  the  product 
of  the  exponent,  the  function  with  its  exponent  diminished  by 
unity,  and  the  first  derivative  of  the  function. 

r 

If  f(x)  =  v  =  u%  where  r  and  s  are  finite,  positive  integers, 
vs  =  ur.     Forming  the  first  derivatives  of  both  sides  of  this 

equation,  s  •  v8-1  —  =  r  •  ur~l  •  — •      Solving   for  — ,   there   re- 
cto dx  dx 

u    dv     r  ur~l  du     r  ur~x  du     r    7-1   du      „ 

suits  —  = ; —  = =  -•?**     Hence 

dx     s   v'~l  dx      s     r-t  dx     a  dx 

d     7     r  JJ-i   du 

—  •  u"  =  -  •  ug ; 

dx  s  dx 


DIFFERENTIATION  AND  INTEGRATION  25 

that  is,  the  first  derivative  of  a  function  affected  by  a  finite, 
positive,  fractional  exponent  is  the  product  of  the  exponent, 
the  function  with  its  exponent  diminished  by  unity,  and  the 
first  derivative  of  the  function. 

If  f(x)  =  u~n,  where  n  is  finite  and  either  integral  or  frac- 
tional, f(x)  =  — ,  and 


dxJ  w      dx\unJ 


du 

dx  __  ,  du 

—  =  —  n  •  u  n    •  — 
dx 


Hence  the  first  derivative  of  a  continuous  function  affected 
by  any  finite  constant  exponent  is  the  product  of  the  expo- 
nent, the  function  with  its  exponent  diminished  by  unity,  and 
the  first  derivative  of  the  function. 

If  u  =  x,  —  xn  =  n  •  xn~l. 
dx 

If  y  is  a  continuous  function  of  x,  x  is  also  a  continuous 
function  of  y.     From  the  equation   y  =fl(x)  an  equation  of 

the  form  x  =  f2(y)  is  obtained.     Differentiation  gives  -^  and 

dx  ...  ^x 

— .     The  relation  between  these  derivatives  is  to  be  found. 

The  equation  — =  1  is  true  for  all  values  of  Ax.  hence 

*  Ax  Ay 


t     .,  A?y  Ax      v     .,  A?/  v     ..Ax      dy  dx      ^ 

limit  — =  limit—  •  limit —  as  -£ =  1 

Ax  Ay  Ax  Ay      dx  dy 


when  limit  Ax  =  0  and  limit  Ay  =  0.     There  results  —  =  —  ; 

dx     dx 

dy 
that  is,  the  first  derivative  of  y  with  respect  to  x  is  the  recip- 
rocal of  the  first  derivative  of  x  with  respect  to  y. 

If  y  is  a  continous  function  of  z,  y  =fx  (z),  and  z  is  a  contin- 
uous function  of  x,  z  =f2(x),  y  is  also  a  continuous  function  of 

x.     The  derivative  -^  is  to  be  calculated. 
dx 


26  DIFFERENTIAL   AND  INTEGRAL   CALCULUS 

The  equation  — *  =  — £  •  —    is  true   for   all  values   of  Ax. 
Ax      Az     Ax 

When  limit   Ax  =  0,   limit   ^  =  limit  ^-  limit  —  :  that  is, 
,  ,       7  Ax  Az  Ax 

dy  _  dy_    dz_^ 

dx       dz     dx 

The  rules  of  this  article  are  sufficient  for  the  differentiation 

of  all  algebraic  functions. 

Example   I.  —  Form  the  first  derivative  of 

5x*-7x2  +  12x-15. 

~  (5x?  -  lx2  +  12  a?  -  15)  =  5  —  a?  -  7 -^-x2  +  12-^-x--^-15 
dx  dx  dx  dx         dx 

=  15x2-Ux-\-12. 


Example    II.  —  Form   the   first   derivative   of   (2  —  5  x2)*. 
This  expression  has  the  form  un  whose  derivative  is 

dx  dx 

In  the  problem  u  =  2  —  5  x2,   n  =  f .     Hence 

A (2  -  5b2)1  =  f(2  -  5^  -|-(2  -  5a?2)  =-  15a?(2  -5a2)*. 

(XX  (XX 

Example  III.  —  Form  the  first  derivative  ®  of  the  implicit 
function  x2  +  y2  =  9.  dx 

Forming  the  first  derivative  of  both  sides  of  this  equation, 

2x  +  2y%>=<>,  whence    &=-*   and  *-=_2. 

dx  dx         y  dy         x 

Example  IV.  —  If  ^  = - —    and   x2  =  z2  +  1,  form  ^L 

<fe      (2»  +  i)f  ax 

From  x2  =  z2  +  1,  —  =  -  •    Hence 


DIFFERENTIATION  AND  INTEGRATION  27 

dy  _dy  dz  _        z*        x _     z2  -x     _  (xP  —  Vj-x _        1 
dx     dz  dx     /z2  _j_  -j\|  z      /z2  i  ^\  §  x2  x 

Example  V.  —  Form  the  first  derivative   ofy  = ~  x    . 

(1  +  »*)* 
Applying  the  rule  for  differentiating  the  quotient  of  two  func- 
tions, 

(1  +  rf.l(l  _  aj)  -  (1  -  a>)  —  (1  +  x2)* 
dy  dxK  J      K  )  dxK    _     ; 

dx  1  +  x2 

-  (1  +  s2)*  -  a;  (1  -  ap  (1  +  x2)~^ 

1  +  x2 

—  1—  X2  —  x-\-x2_  1+x 


(1  +  x2)*  (1  +  «0j 


PROBLEMS 
Form  the  first  derivative  of, 

*.»•  +  &  6.   -L-. r  12.   (3  +  5x)"i. 

1  —  ar  ! 

2.  2a2  +  7a  +  3.  1  +  a2  13*    (fX^)  * 

'    1  —  x2 

3.  3  a* -8.  „        N9  14.    (5a; -7a2)1. 

8.    (1  —  x)2. 

4.  -A—  9.  a-x2)2.  15-  (i-«  +  aOf- 

1  +  x.  K  J 

10.    (l-a;)~2.  16.    (a  +  bxn)m. 

5.  i±*  .  - 

1  —  x  11.    (1  —  a;2)*.  17.    (a  +  &af)tt. 

±  —  x2  2 

18.    Form    the    first   derivatives    of and  -,  and 

1  _i_  x^  1   \  x 

find  the  difference  of  the  functions. 


28  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

In  the  following  equations  y  is  an  implicit  function  of  x. 

Form  dX 
dx 

a2     b2  a2     b2 

22.    x2y-\-y  —  a;  =  0.  23.    a?  -  3xy  +  f  =  0. 

Determine  the  rate  of  change  of  function  and  variable  of 

24.  f(x)  =  2  x  -  x2.  26.  fix)  =  (1  -  a2)*  at  *=  f 

25.  f{x)  =  T-—J  27.   /(*)  =  (4  -  x2^  at  *  =2. 

28.  Find  at  what  point  of  the  circle  x2  +  y2  =  9  the  ordinate 
increases  twice  as  fast  as  the  abscissa. 

29.  Discuss  the  rate  of  change  of  ordinate  and  abscissa  of 
^4-^=1  for  different  points  of  the  ellipse. 

The  rate  of  change  of  ordinate  and  abscissa  is  measured  by 

-S5L= When  x  and  y  have  like  signs  the  ordinate  is  a 

dx  dry 

decreasing  function  of  x;  when  x  and  y  have  unlike  signs  the 

ordinate  is  an  increasing  function  of  x ;  when  x  =  0?  the  ordi- 
nate is  not  changing  value  ;  when  y  =  0,  the  ordinate  changes 
infinitely  more  rapidly  than  the  abscissa.  These  results  agree 
with  the  results  obtained  by  examining  the  ellipse. 

30.  Determine  the  rate  of  change  of  area  and  side  of  an 

V5 
equilateral  triangle.     Area  =  -—-  x2,  x  being  a  side. 

31.  Find  the  rate  of  change  of  area  and  radius  of  a  circle 
when  the  radius  is  10. 

32.  A  man  walks  on  level  ground  towards  a  tower  80  feet 
high.  When  60  feet  from  the  foot  of  the  tower  find  the  rela- 
tive rate  of  approaching  the  top  and  foot  of  the  tower. 


DIFFERENTIATION  AND  INTEGRATION  29 

Calling  the  man's  distance  from  the  top  of  the  tower  y,  his 
distance  from  the  foot  of  the  tower  x,  y2  =  x2  +  6400. 

Find  the  slopes  of  the  tangents  to  the  following  curves  at 
the  point  (x,  y)  of  the  curve : 

2  2 

33.    y  =  ±x2.      34.    y  =  6x-x2.       35.    —--L  =  l. 

a2      b2 

36.  Find  the  slope  of  the  tangent  to  y  =  8  x  —  x2  at  x  =  1. 

37.  Find  where  the  point  generating  the  circle  x2  -f  y2  =  4 
tends  to  move  parallel  to  the  X-axis.  Where  parallel  to  the 
F-axis. 

Determine  the  velocity  at  time  t  supposing  the  relation 
between  s  the  distance  and  t  the  time  to  be  expressed  by  the 
folloAving  equations,  where  a  and  u  are  constants : 

38.  s  =  £  at2.      39.    s  =  ut  +  \  at2.      40.    s  =  ut  —  \  at2. 

41.  If  ^  =  z2(l  +  z)\  and  1  +  z  =  x2,  find  & 

dz  dx 

42.  If  ^  = and  1  +  z  =  x*r  find  JL 

d*      VI  +  z  d>x 

Art.  13.  —  Integration 

The  difference  between  two  functions  which  have  the  same 
first  derivative  is  constant.  Hence,  if  the  first  derivative  of  a 
function  is  known,  the  function  itself  is  known,  except  for  an 
additive  arbitrary  constant.  The  process  of  obtaining  a  func- 
tion f(x)  from  its  first  derivative  — f(x)  is  called  integration, 

/ax 
The  operations  denoted  by 

the  symbols  —  and  f  neutralize  each  other  ;  that  is, 
f±m-fto+0    and    £/>(*)=/(*), 


30  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

where  C  is  an  arbitrary  constant.  Hence  the  rules  of  integra- 
tion are  at  once  inferred  from  the  rules  of  differentiation. 
The  result  of  integration  is  called  an  integral,  and  the  integral 
is  correct  if  the  derivative  of  the  integral  is  the  expression 
which  was  integrated. 

The  rules  of  algebraic  integration,  obtained  from  the  rules 
of  algebraic  differentiation  of  Art.  12,  are  expressed  by  the 
formulas : 

i.  C^  =  u+a  in.  Ci  =  x+a 

J  dx  J 

II.    fo  =  <7.  IV.    Ca.  —  =  a>u+C. 

J  J      dx 

•s     V  UjJU  UjJLi         llJU 


VI      Cf    .  —  4-     .— 

J  \     dx  dx 


du  dv 

V u 

_____  \X*Ju  LtJO 


u*v  +C. 


/ax  dx     u  .  n 

S»  ~v+a 

VIII.    f un  •  —  =  -^1  -f-  G.    When  u  =  x,  this  formula  be- 

J  dx       71  +  1 

/xn+1 
xn  — 4-  C.      When    n  =  —  1,    this    result    is 
n  +  1 


comes 
absurd. 


The  rule  of  formula  VIII.  may  be  stated,  the  integral  of  any 
function  affected  by  an  exponent  other  than  —  1,  multiplied 
by  the  first  derivative  of  the  function,  is  the  function  with  its 
exponent  increased  by  unity  divided  by  the  increased  expo- 
nent, plus  an  arbitrary  constant. 


DIFFERENTIATION  AND  INTEGRATION  31 

Since  la-  —  =  a-u+C  and   a  I  —  =  a-w+Cf,  a  constant 
J       dx  J  dx 

factor  may  be  shifted  from  one  side  of  the  sign  of  integration 
to  the  other,  without  affecting  the  result. 

By  formula  V.  the  integral  of  the  algebraic  sum  of  a  finite 
number  of  functions  is  the  like  algebraic  sum  of  the  integrals 
of  the  functions. 

Example  I.  —  Find  the  function  of  x  whose  first  derivative 
is  4  a2  —  5 x.    Denoting  the  function  by  f(x),  — f(x)  =  4  a?2  —  5a? 

and  f(x)  =  I  (4a;2  —  5x)  =  A  I  x2  —  5  I  x  =  ^x?—  fa2  +  C,  where 

the  arbitrary  constant  C  is  called  the  constant  of  integration. 
This  integral  is  called  the  indefinite  integral.  If  f(x)  is  re- 
quired to  have  a  given  value  for  a  given  value  of  x,  for 
example,  if  f(x)  =  10  when  x  =  1,  the  value  of  C  is  found 
from  the  equation  f(x)  =|a^  —  f  x2  +  C  to  be  Hi.  Hence 
under  the  given  conditions  f(x)=^x3  —  ^x2  +  ll^.  This  re- 
sult is  called  the  corrected  integral. 

Example  II.  —  Find  the  function  of  x  whose  first  derivative 
is  x  (1  +  x2)^.     Denoting  the  integral  by  f(x), 

f(z)=fx(l +  «?)*. 

This  integration  can  be  performed  by  means  of  the  formula 

/un  -  —  =  — - — -  +  C  if  x  (1  +  x2y  can  be  separated  into  factors 
dx     n  +  1 

of  the  form  un  and  — .     Placing  u  =  1  +  x2 ,  —  =  2  x.     Hence 
dx  dx 


32.        DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Example  III.  —  If  ^  =  10  (x  -  2)  (a*  -  4  x)~l}  find  y. 
ctx 

Placing  u  =  x2-±x,  —  =  2x-4, 
da; 

and         2/  =  10  C(x  -  2)  (a2  -  4  a)""* 

=  5  f(z2_4a;)-|— (a;2-4a;)  =  10  (a2  -  4  a)i -f  01 
*/  da; 

Example  IV.  —  Determine  the  equation  of  a  curve  such  that 
the  slope  of  the  tangent  to  the  curve  at  any  point  (x,  y)  is  the 
negative  ratio  of  the  abscissa  to  the  ordinate. 

The  condition  of  the  problem  is  expressed  by  the  equation 

-^  =  —  -,  whence  y^-  =  —  x.  Integrating,  -y2  =  — -x2  -\-  C  or 
dx         y  dx  2  2 

x2  +  y2  =  2  C.  This  equation  represents  any  circle  with  center 
at  the  origin  of  coordinates.  If  the  circle  is  required  to  con- 
tain the  point  (3,  4),  the  value  of  2  C  must  be  25,  and  the 
problem  has  the  determinate  solution  x2  +  y2  =  25, 

PROBLEMS 

Find  the  f(x)  whose  first  derivative  is, 

1 .  2  +  x,  knowing  that  f(x)  =  7  when  x  =  2. 

2.  3  —  5  x2,  knowing  that  f(x)  =  20  when  x  =  5. 

3.  1  +  x  +  x2  +  Xs,  knowing  that  f(x)  =  12  when  x  —  1. 

4.  aA  5.   3afi  6.   x%  +  5.  7.   2a*  —  aA 

8.  (1  —  #)  (2  +  x2).      Multiply  out  and  integrate  term  by 
term. 

9.  (3z-5)(2a-3a;2).  12.    (2 s  +  3)  (x2  +  3 x)i 

10.  (1  +  a2)  (3ar>  +  2).  13.    (3a2  -  10a?)  (x3  -  5^)1 

11.  (4a;-5)(2a;2-5a;)l  14.    (2x  +  5)(x2  +  5x -7)i 


DIFFERENTIATION  AND  INTEGRATION 


33 


15.  (l-f9a?)f  17.  15  (3  x2  -8  x)  (3 a? -12  a?8)* 

16.  (a  +  fcu)*.  18.    (a2  +  3  as  -5)  (2  a? +  9  a2  -30  a)  I 

Find  the  equation  y=f(x)  of  a  curve  such  that  the  slope 
of  the  tangent  at  any  point  (a,  y)  is, 

19.  3x  —  7. 

20.  2  a  +  5,  the  curve  passing  through  (5,  0). 

21.  a^-f  5  a?,  the  curve  passing  through  (0,  0). 

b2  x 

22.  - ■  •-,  the  curve  passing  through  (a,  0). 
a2  y 

23.  For  the  cable  of  a  suspension  bridge  with  load  uni- 
formly   distributed     over    the 


horizontal, 


dy 


—  'X,  where  w 


1 

1 

J&F^ 

y 

Fig.  0. 


dx  vQ 
is  the  uniform  load  per  hori- 
zontal linear  unit  and  t0  is  the 
tension  at  the  lowest  point. 
Find  equation  of  curve  assumed 
by  cable. 

24.    Find  the  function  of  x  whose  rate  of  change  is  2x  —  5. 

Find  the  relation  between  s  and  t,  knowing  that  s  =  0  when 
t  =  0,  and  that  the  velocity  is, 

25.    u-\-at.  26.    u  —  at. 

Find  the  relation  between  x  and  y,  knowing  that, 
07     dy_  —  ®!_  oa     <%_!— 3a-f5a2 

dx      y  dx  y  —  ^y3 

28.    32~i±iL  31.    ^  =  JL. 

cfa     1  —  2/  da?     a2  2/3 


29.    3SU-*. 

da     1  +  y2 


32.   ar 


^  =  5 

da     2/ 


33.      f       *        ■  34.      f £— r  35.      f(l-^)3. 


36. 


34  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Find  the  values  of  the  following  integrals : 

:  *   .     34.  r   a  s 

Vl  -  x2  J  (1  -  ^)f 

f(l±^)\  37.      f-  -. 

J\XJ  J  (2ax-  x2)* 

Multiplying  numerator    and  denominator   by   (x~2)~2 =x~5, 
the  problem  becomes 

f — =  -  i  f (2  ax-1  -  l)-f— (2  ax-1  -  1) 

=  -  (2  ax-1  -  l)-i  +  C  = +  C. 

a  a  V2  ax  —  x2 

38<  r V2  ax  -  x2.     39    r — i —     40#  r    g   . 

■J  x3  ^  (a2  +  x2)*  ^  Vl  +  x 

Calling  the  integral  y,  the  problem  is,  given  -^,  to  find  y. 

ax 

By  the  substitution  1  +  x  =  z2, 

dy  =  dy  dx=       x       ,oz  =  2z^  ~^  =2z2 -2. 

dz      dx  dz      Vl+~a;  * 

o 
By  integration  y  =  vz*  —  2z  +  C, 

hence  y  =  |(1  +  x)f  -  2(1  +  x)i  +  C. 

41.      fx2(l+x)l  42.      f — -• 

J  J  (x2  +  1)! 

Art.  14.  —  Definite  Integrals 

If  J- *»  =/(»),    j7(x)  =  F{x)  +  C      If    *\x)  +  G    is 

evaluated  for  x  =  6  and  x  =  a  and  the  second  result  subtracted 
from  the  first,  the  final  result  F(b)  —  F(a)  is  called  the  defi- 


DIFFERENTIATION  AND   INTEGRATION  35 

nite  integral  of  f(x)  between  the  limits  x  =  a  and  x  =  b.    This 
operation  is  denoted  by  writing 


J7<*)=[* 


F(x)  +  C 


=  F(b)-F(a), 


where  a  is  called  the  lower  limit  and  b  the  upper  limit  of  the 
definite  integral. 

From  this  definition  it  follows  that 

and  also  that,  if  &  —  a  =  (b  —  c)  +  (c  —  a), 

m=jc /(*)+)/(*)■ 

Example.  —  Find  the  value  of  the  definite  integral 
f5(3  x2  -  5  x  +  7). 

C\3x2-5x  +  7)=[x'i-%x2  +  7x+cY~  =63. 

PROBLEMS 

Calculate  the  definite  integrals, 
1.     f2(3^-5).  6.     C2x(4,-x2)K 


2.     f5(2x2H-4x-6). 

«/2 

7.      Cx(l-x2)?. 

3.    j'+1^(4-orJ). 

8.     fa  (a2 -a2)*. 

4.  ^4(i  +  ^)a-^2)- 

9.     J    a^cfaj. 

5.    Jfd-rf). 

/•«           2 

10.     1    3x*dx. 

36  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

11.  fVs/l  +  a.     Write  -s^VT+J  and  determine  ^ 
Jo  dx         ,  ,       ,  e?z 

when  1  +  m.  =  z2.       There    results  —  =  —  •  —  =  2  22(z2  -  l)2. 

dz       dx    dz 

Since  z  =  l  when  x  =  0,  and  2  =  2  when  x  =  3, 

u  =  C*a?y/T+x  =  f22  z2(z2  -  I)2  =  16.1. 

12.  If  s  is  the  distance  in  feet  and  t  the  time  in  seconds  of 
a  body's  motion,  find  the  distance  the  body  moves  from  the 
end  of  the  third  to  the  end  of  the  fifth  second,  knowing  that 

—  =  32.16*. 
dt 

Art.  15.  —  Evaluation  of  the  Limit  of  the  Sum 

x  =  b 

The  value  of  S  =  limit  2/(x)  •  Ax  when  limit  Ax  =  0,  where 

x  =  b  x  =  a 

2/(#)  •  Ax  stands  for 

x  =  a 

f(a)  •  Ax+f(a  +  Ax)  -  Ax-] hf(p— 2  Ax)  •  Ax+f(b  —  Ax)  •  Ax, 

is  to  be  determined.  Suppose  F(x)  to  represent  a  continuous  func- 
tion of  x  whose  first  derivative  is  f(x),  that  is,  — F(x)  =  /(x). 
This  means  that — ^— — — -^JZ — Li=y(#)^Cj  where  e  be- 
comes indefinitely  small  when  Ax  becomes  indefinitely  small. 
From  the  last  equation,  /(x)  •  Ax  =  F(x  -\-  Ax)  —  F(x)  ±  c  •  Ax. 
Whence  by  substituting  for  x  successively  a,  a  +  Ax,  a  +  2  Ax, 
•••,  ft  —  2  Ax,  b  —  Ax, 

/(«)  •  Ax  =  F(a  +  Ax)  —F(a)±£1-  Ax, 

f(a  +  Ax)  •  Ax  as  F(a  +  2  Ax)  —  F(a  +  Ax)  ±  c2  •  Ax, 

/(a  +  2  Ax)  •  Ax  =  F(a  +  3  Ax)  -  F(a  +  2  Ax)  ±  e3-  Ax, 

f(b  -  2  Ax)  •  Ax  =  JF(6  -  Ax)  -  F(b  -  2  Ax)  ±  en_x  .  Ax, 
f(b-Ax)-Ax    =F(b)-F(b-Ax)±en.Ax. 


DIFFERENTIATION  AND  INTEGRATION  37 

x  =  l 

By  addition,  %f(x)  •  Ax  =  F(b)  -  F(a)  ±  2e  •  Ax.     Since  F(x) 

x  =  a 

is  continuous,  each  c  becomes  indefinitely  small  when  Ax  be- 
comes indefinitely  small.  Denoting  by  E  the  numerically 
largest  e,  %e-Ax  ^E"XAx  =  E(b —  a),  which  becomes  indefi- 
nitely small  when  E  becomes  indefinitely  small,  since  b  —  a  is 
supposed  to  be  finite.     Hence 

8  =  limit  %f(x)  •  Ax  =  F(b)  -  F(a), 

x=-a 

when  limit  Ax  =  0.     But  F(b)  —  F(a)  is  the  value  of  the  defi- 

f(x),  for  by  hypothesis  — F(x)=f(x).   Hence, 
«  dx 

finally,  if  ^-F(x)=f(x),  S  =  limit'I /(») Ax  =  f/(«)  when 

limit  A#  =  0. 

If  the  summation  limits  are  a  and  #,  the  result  obtained 
becomes  S=   J  f(x)  =F(x)  —  F(a).     Forming  the  derivative 

oiS,—  =  —F(x)  =  f(x). 
9  dx      dx     w     Jy  J 

Art.  16.  —  Infinitesimals  and  Differentials 

A  quantity  which  becomes  indefinitely  small  is  called  an 
infinitesimal. 

If  several  infinitesimals  occur  in  the  same  problem,  any  one 
may  be  chosen  as  the  principal  infinitesimal.  Denote  the 
principal  infinitesimal  by  a. 

Infinitesimals  whose  ratio  to  a  is  finite  are  said  to  be  of 
the  first  order. 

An  infinitesimal  /3  whose  ratio  to  the  nth  power  of  a  is 
finite  is  said  to  be  of  the  nth  order.  If  n  is  positive  and 
larger  than  unity,  ft  is  said  to  be  of  a  higher  order  than  a. 

Denoting  the  finite  ratio  of  B  to  an  by  r,  -£  =  £ r=fi 

an      a  an~l 


38  DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

whence  ^=?-.«r,_1,  an  infinitesimal,  and  -  = -,  an  inden- 
ts                                                        fi      r-a11'1 

nitely  large  quantity.  Hence,  if  /?  is  an  infinitesimal  of  a 
higher  order  than  a,  /?  =  e-  a,  where  e  is  an  infinitesimal. 

The  laws  governing  the  use  of  infinitesimals  are  contained 
in  the  following  two  propositions. 

I.  In  the  limit  of  the  ratio  of  infinitesimals  any  infinitesimal 
may  be  replaced  by  another  infinitesimal  differing  from  it  by 
infinitesimals  of  higher  orders. 

Let  a  and  p  represent  any  two  infinitesimals,  and  consider 
the  ratio 

r ~  A,-p  +  Br  /?2  +  Cr  P3  +  A-  P4  +  •- 
a    A  +  B-a+C'a2  +  D'a3  +  -> 

where  the  coefficients  B,  C,  D,  •••  and  Blf  Clf  A?  •••  are  finite. 
Let  M  be  the  numerically  largest  of  the  coefficients  B,  C,D,---, 
My  the  numerically  largest  of  the  coefficients  B1}  Clf  A?  •••• 
Then  B .  a  +  C>  «2+  D  •  a3  +  —>M(a  +  a?  +  a3+  ».)  =M 
an  infinitesimal,  and 
Brp+Crp>2  +  D1.p3+:.>M1(p  +  p2  +  p3+>..)=Mir-tL 


l-V 


1-/*' 

another  infinitesimal.     Hence  in  the  limit 

a     A  +  B-a  +  (7- a2  +  #•  <*3  +  ♦••    _Aa 
r~~pA1  +  Bl.p+Cl>'p  +  D1.IP+..>      Bp 

II.  In  the  limit  of  the  sum  of  infinitesimals,  provided  this 
limit  is  finite,  any  infinitesimal  may  be  replaced  by  another 
differing  from  it  by  an  infinitesimal  of  a  higher  order. 

Suppose  limit  fa  +  a2  +  «3  -\ )  =  c,  a  finite  quantity,  and 

let  ft  =  ai  +  «i  •  «i,  ft  —  «2  +  €2  •  «2,  ft  =  «3  +  e3  •  a3,  •••,  where 


DIFFERENTIATION  AND  INTEGRATION  39 

ci>  c2>  c3>  »••  are  infinitesimals  of  which  e  is  the  numerically 
largest.     Adding 

=  («1  +  «2  +  <*3  H )   +  («1  *  «1  +  C2  '  «2  +  C3  *  «3  +    •••)• 

Now  c1.«1  +  e2.«,  +  e3.a34-'->«(«i+«2  +  «2H )  =  e  •  e,  an 

infinitesimal.     Hence  in  the  limit 

ft  4-  ft  +  ft  +  -  =  «i  +  «2  +  *»  +  •••• 

This  proposition  is  true  even  when  c  is  indefinitely  large,  pro- 
vided e  •  c  is  an  infinitesimal. 

If  the  limit  of  the  ratio  of  two  infinitesimals  is  unity,  the 
infinitesimals  can  differ  only  by  infinitesimals  of  higher  orders. 
For  if  limit  -  =  1,  —  =  1  ±  e  and  B  =  a  ±  e  •  a,  where  c  is  an 

infinitesimal.  Hence  the  rules  governing  the  use  of  infinitesi- 
mals may  be  stated,  in  the  limit  of  the  ratio  and  in  the  limit 
of  the  sum  any  infinitesimal  may  be  replaced  by  another 
infinitesimal,  provided  the  limit  of  the  ratio  of  the  two  infini- 
tesimals is  unity. 

Since  infinitesimals  of  higher  orders  disappear  from  the 
limit  of  the  ratio  and  from  the  limit  of  the  sum,  the  solution 
of  problems  involving  the  limit  of  the  ratio  or  the  limit  of  the 
sum  may  be  simplified  by  dropping  infinitesimals  of  higher 
orders  at  the  start. 

If  y=f(x)  is  a  continuous  function  of  x  whose  first  deriva- 
tive is  /'<c,  — ^  ==  f  (x)  -f  e,  whence  Ay  =  f'x  •  Ax  +  c  •  Ax,  where 
Ax 

Ay  is  the  difference  in  the  value  of  the  function  corresponding 
to  a  difference  of  Ax  in  the  value  of  the  variable,  and  c  be- 
comes an  infinitesimal  when  Ax  becomes  an  infinitesimal. 
When  the  difference  Ax  becomes  an  infinitesimal  it  is  denoted 
by  dx,  which  is  read  differential  x.     The  change  in  the  value 


40  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

of  y  corresponding  to  dx  is  f  (x)  •  dx  -\-e-dx.  Defining  dy, 
read  differential  y,  by  the  equation  dy  =/'  (a?)  •  dx,  dy  differs 
from  the  change  in  the  value  of  y  corresponding  to  a  change 
of  dx  in  the  value  of  a;  by  e  •  dx,  an  infinitesimal  of  a  higher 
order  than  dx.  Hence  in  problems  involving  the  limit  of  the 
ratio  or  the  limit  of  the  sum  the  actual  change  in  y  may  be 
replaced  by  dy  =/'  (x)  •  dx. 

The  first  derivative  of  a  function  is  therefore  the  factor  by 
which  the  differential  of  the  variable  must  be  multiplied  to 
obtain  the  corresponding  differential  of  the  function.  This 
explains  why  the  first  derivative  of  a  function  is  frequently 
called  the  differential  coefficient  of  the  function. 

The  operation  of  finding  the  differential  of  a  function  corre- 
sponding to  the  differential  of  the  variable  is  called  differentia- 
tion. It  will  be  observed  that  differentiation  as  here  defined 
is  performed  by  the  rules  established  for  differentiation  as 
defined  in  Art.  12. 

For  example,  if  y  =  Xs  —  7  x2  -f  15  x  -f  10, 

dy  =  j-  (x3  -  7  x2  +  15  x  +  10)  dx  =  (3  x2  -  14  x  +  15)  dx. 

If  F(x)  is  a  continuous  function  of  x  whose  first  derivative 
is  f(x),  &F(x)  =  F(x  +  Aaj)  —  F(x)  =f(x)  •  Ax  +  e  •  Aaj,  where 
AF(x)  and  c  become  infinitesimals  when  A#  becomes  an  infini- 
tesimal.    The  sum  of  the  elements  AF(x)  of  F(x)  from  x  =  a 

x=x 

to  x  =  x  is  2^(aj)  —  F(a)  =  2  [/(»  .  Aa?  +  c .  Aaj],  and  the  sum 

z  =  a 

of  the  elements  Aa;  of  x  from  x  =  a  to  a;  =x  is  a?  —  a.  This 
is  true  for  all  magnitudes  of  Aa;.  When  the  element  Aa; 
becomes  the  infinitesimal  element  dx, 

F(x)  -  F(a)  =  i\f(x)  .&-f£.  dx]. 


DIFFERENTIATION  AND  INTEGRATION  41 

Supposing  x  —  a  to  be  finite,  in  the  limit 

F(x)  -  F(a)  ="?/(*)  •  dx, 

x  =  a 

since  e-dx  is  an  infinitesimal  of  a  higher  order  than  dx. 
Calling  the  operation  of  finding  the  limit  of  the  sum  integra- 
tion and  denoting  it  by  the  symbol  j ,  and  indicating  that  the 
sum  is  to  extend  from  x  =  a  to  x  =  x  by  writing    |    ,  the  result 

fix)  •  dx  =  F(x)  —  F(a).     Now 
F(x)  —  F(a)  was  found  in  Art.  14  to  be  the  value  of  the  defi- 
f(x),  provided  — F(x)  =f(x). 

a>  dX 

In  general,  if  y=f(x)  and  ^-=f'(x),  in  the  notation  of 

dx  ,, 

differentials  dy=f  (x)  •  dx  and  y  =  (/'(#)•  ^  —f(x)  +  & 
Differentiation  and  integration  as  defined  in  this  article  are 
again  inverse  operations. 

It  will  be  observed  that  integration  as  here  defined  is  per- 
formed by  the  rules  established  for  integration  in  Art.  13. 
For  example,  if 

dy=  (3x2-5x  +  l0)dx,  y  =  X*  -fa;2  +  10a; -f  O; 

the  sum  of  the  elements  of  y  from  x  =  0  to  x  =  x  is 

CiSx2  -  5  x  +  10)  dx  =  [a?  -  fa;2  +  10a;  +  C~]'q 
=x3_fa;2  +  10a,. 

the  sum  of  the  elements  of  y  from  x  —  0  to  x  =  4  is 

f4(3  a^  -  5  a;  +  10)  dx  =  [a;3  -  f  x2  +  10  a;  +  c]  ]  =  64. 

While  the  method  of  differentials  is  based  on  the  method  of 
limits,  the  method  of  differentials  has  two  decided  advantages  : 
first,  calculations  are  simplified  by  dropping  differentials  of 
higher  order  at  the  start ;  secondly,  the  successive  steps  in  the 


42         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

application  of  the  method  of  differentials,  especially  in  sum- 
mation problems,  are  more  directly  intelligible  than  is  the  case 
when  using  the  method  of  limits.  Hereafter  the  derivative 
and  differential  notations  are  used  interchangeably. 

PROBLEMS 

Form  the  differentials  of  the  functions, 

x2-! 


1.  y  =  Sx2  —  5.  3.   y  = 


x2  +  l 


2.   y=(l  — a2)*.  4.    y= — 

Find  the  functions  whose  differentials  are, 

5.   dy=(2xi-5x)dx.  6.    dy  =  (1  +  x2)  dx. 

7.   dy=(l  +  xydx. 

Evaluate  the  definite  integrals, 

8.     C(4:X-5)dx.  9.     C\2xi-6x)dx. 

10.     C\x2-3x  +  5)dx. 

11 .  Cx*  (x2  +  1)~*  dx.      Placing    x2  + 1  =  z2,  dx  =  -  dx,  and 

xs(x2  +  l)~^dx  =  xZ'Z-3'~'dz  =  x2'Z-2-dz 

X 

=  (z2-l)-z-2-dz  =  dz-^ 

v         }  z2 

While  x  takes  all  values  from  0  to  3,  z  takes  all  values  from  1 
to  10.     Hence 

£  V  (x2 + iy-i  dx=  £°fdz  -  £?\ = r9  +|  +  cT° = 8.i. 

12.  C  a?(l  +  x)ldx.     - 


CHAPTER  IV 

APPLICATIONS   OF  ALGEBRAIC   DIFFERENTIATION  AND 
INTEGRATION 

Art.  17.  —  Tangents  and  Normals 


The  slope  of  the  tangent  to  the  curve  whose  equation  is 
y  =f(x)  at  any  point  (x0,  y0)  is  the  first  derivative  of  y  =  f(x) 
evaluated   for   x  =  xQ,    y  =  y0.      This    is   denoted   by   writing 

tana  =  -^.     The  equation  of  the  tangent  TT'  to  y=f(x)  at 
dx0  , 

(a?0,  y0)  is  y-y0  =  Jh(x-Xo). 
ax0 

The  intercept  of  the  tangent 
on  the  X-axis,  found  by 
placing  y  =  0  in  the  equa- 
tion of  the  tangent  and  solv- 
ing for  x,  is  AT=  x0— y©~?; 
the  intercept  on  the   Y-axis 


is  AT'  =  y«  —  x0 


dx0 


The  portion  of  the  tangent  FlQ  7 

bounded  by  the  point  of  tan- 

gency  and  the  point  of  intersection  of  the  tangent  with  the 
X-axis  is  called  the  length  of  the  tangent.     From  the  figure 

43 


44  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


the  length  of  the  tangent  is  PT  =  y0yjl  +  f |&Y  The  projec- 
tion of  the  tangent  PT  on  the  X-axis  is  called  the  subtangent. 

dx 
From  the  figure  the  subtangent  is  Z>jT  =  2/0— °- 

dy0 
The  equation  of  the  normal  NN'  to  y=-.f(x)  at  (a*,,  y0)  is 

y  —  y0  =  —  =5 ?  (x  —  x0).     The  intercept  of  the  normal  on  the 

X-axis  is  AN=x0  +  y0-^',   tne  intercept  of  the   normal   on 
dx0 

the  F-axis  is  AN'  =  2/o  +  ^o-r9. 

The  portion  of  the  normal  bounded  by  the  point  (x0,  y0)  and 
the  intersection  of  the  normal  with  the  X-axis  is  called  the 
length  of  the  normal.     From  the  figure  the   length  of  the 

normal   is  PN  =  2/0  \/l  -f  [  —  ] .     The  projection  of  PN  on 
the  X-axis  is  called  the  subnormal.     From  the  figure  the  sub- 
normal is  DN=  2/0— • 
dx0 

Example.  —  Find  the  equations  of  tangent  and  normal  to 
x2  -\-2y2  —  2xy  —  x  =  0  at  the  point  (1,  1).  Also  the  length 
of  tangent  and  subtangent,  and  of  normal  and  subnormal. 
Differentiating    x2  +  2y2  —  2  xy  —  x  =  0   with    respect    to    x, 

2*-My&_2y-2i3Ul«0,  whence  *SL  =  l±M=M 
dx  dx  dx         4  2/  —  2x 

At  the  point  of  tangency  x0  =  1,   y0  =  1,  ^  =  -,    — °  =  2. 

dx0      2     dy0 

Hence  the  equation  of  the  tangent  is  y  —  1  =  \{x  —  1),  reducing 
to  y  =  ^x  +  £ ;  the  equation  of  the  normal  y  —  1  =  —  2  (a;  —  1), 
reducing  to  y  =  —  2  x  +  3 ;  the  length  of  the  tangent  is  V5  > 
the  length  of  the  subtangent  2;  the  length  of  the  normal 
^  V5 ;  the  length  of  the  subnormal  ^- 


APPLICATIONS  45 


PROBLEMS 


5.   Find  the  subtangent  of  the  ellipse  —  +  —2=  1. 


1.  Find  the  equations  of  tangents  and  normals  to 
x2y  +  y  —  x  =  0  at  x  =  -\-l  and  at  x  =  —  1. 

2.  Find  the  equations  of  tangent   and   normal   to  xy  =  4 

at  x  =  2. 

3.  Find  equation  of  tangent  to  x*  +  y*  —  a¥  at  (x0,  y0). 

4.  Show  that  the  sum  of  the  intercepts  of  the  tangent  to 
x^  +  y^  =  a*  is  the  same  for  all  positions  of  the  point  of 
tangency. 

6.  Find  the  subnormal  of  the  parabola  y2  =  2px. 

7.  Find  the  length  of  the  normal  to  4  x2  +  16  y2  =  100  at 
x  =  3. 

8.  Find  the  length  of  the  tangent  to  4ar  -f- 16/  =  100  at 
x  =  3. 

9.  Find  the  length  of  the  subnormal  to  x2  +  y2  =  25  at 
(3,  4). 

10.   Determine  the  curves  whose  subnormal  is  constant. 

The  hypothesis  is  expressed  by  the  equation  y—  =  a,  where 
a  is  the  constant  length  of  the  subtangent.  From  this  equa- 
tion —  =  2    or    adx  =  ydy.       By    integration    ax  =  \y2  -f-  C 

dy      a 
or  y2  =  2  ax  —  2  C.      This   equation  represents   a  system   of 

parabolas. 

Art.  18. — Length  of  a  Plane  Curve 

Denote  by  s  the  length  from  x  =  a  to  x  =  b  of  the  plane 
curve  whose   equation  is   the   continuous   function   y=f(x). 


46 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Divide  ab  into  any  number  of  equal  parts  Ax,  and  draw  ordi- 
nates  at  the  points  of  division  of  ab.      Draw  the  chords  of 

the  arcs  As  into  which   these 
Y  ordinates  divide  the  curve  AB. 

At  the  ends  of  any  one  of  these 
arcs,  such  as  mn,  draw  tangents 
to  the  curve  AB.  Assume  as 
axiomatic  that  chord  mln  <  arc 
mn  <  broken  line  mkn.  Prom 
the  right  triangles  Mm  and  Mn, 

km  =  Im  -  sec  kml, 
kn  =ln  '  sec  knl. 


** 

L       -^ 

R 

m 

AV 

A  x 

A/ 

C 

I 

> 

Fig.  8. 


These  relations  are  true  for  all  magnitudes  of  the  chord  mn. 
As  the  length  of  the  chord  mn  is  diminished,  the  angles  kml 
and  knl  approach  zero  and  their  secants  approach  unity. 
Hence  km  =  Im  (1  -+-  cx)  and  kn  =  In  (1  +  c2),  where  cj  and  e2 
approach  zero  when  mn  approaches  zero.     Adding, 

km  +  kn  =  (Im  +  foi)  —  (cj  •  Zm  +  c2  *  &*)• 

When  the  chord  mln  becomes  an  infinitesimal,  Im,  In,  ely  and  c2 
become  infinitesimals.     It  follows  that 

broken  line  mkn  —  chord  mln  =  ex  •  Im  +  c2  •  Zn, 

an  infinitesimal  of  a  higher  order  than  the  chord  mln.  Hence 
the  difference  between  the  infinitesimal  arc  mn  and  its  chord 
mln  is  an  infinitesimal  of  a  higher  order  than  the  chord,  and 
in  problems  involving  the  limit  of  the  ratio  or  the  limit  of  the 
sum,  the  infinitesimal  arc  may  be  replaced  by  its  chord. 

It  follows  at  once  that  the  length  of  the  curve  is  the  limit 
of  the  length  of  the  inscribed  broken  line  when  the  number  of 


APPLICATIONS  47 

parts  into  which  ab  is  divided  is  indefinitely  increased,  which 
is  equivalent  to  saying  when  limit  Ax  =  0.     That  is, 

s  =  limit'i^Az2  +  Ayrf  =  limitsYl  +  ^Y  •  Ax 

x  =  a  x  =  a\  Ax2) 

when  limit  Ax  =  0. 


Hence  S=   fYl  +  ^-rf*    and    £  =  (l  +  f^ 

J  a  V         doer  J  ax      V         dx 


2M 

dx2)  dx     \     '  dx2; 

Since  -^  =  tan  a,  where  a  is  the  angle  of  inclination  to  the 
dx 

X-axis  of  the  tangent  to  the  curve  y  —f{x)  at  (x,  y), 

/Jo  1  fit* 

—  =  (1  -f  tan2  aY  =  sec  a,  whence  cos  a  =  — • 
dx     v  J  '  ds 

Since  sin  a  =  tan  a  •  cos  a,  s'mu  =  ~  —  =  -^.     These   results, 

dx  ds      ds 

obtained  by  the  method  of  limits,  may  be  roughly  inferred 
from  the  figure.  As  Ax  approaches  zero,  the  element  of  curve 
As  approaches  equality  with  its  chord  and  becomes  a  part  of 
the  tangent.  Hence  in  the  limit,  when  Ax,  Ay,  and  As  become 
the  infinitesimals  dx,  dy,  and  ds, 


=dx2+c¥=(i+|^y^  d,s=(i+ 


<¥Y^    --A.*,- 


,  dy        .  dy  dx 

tan  «  as  -2-3     sin  a  =  — ,     cos  a  =  — 
dx  ds  ds 

Example.  —  Find   the   length   of  the   semi-cubic   parabola 
y2  =  4  x3  from  x  =  5  to  x  =  10. 

From  the  equation  of  the  curve,  y  =  2  x2  and  -^  =  3  x\ 

dx 

Xio  ,  r  .         -iio 

(l  +  9z)*.cfa  =    _2T(l  +  9a^+<7     =31.07. 

The  length  of  the  curve  from  the  origin  (0,  0)  to  any  point 


48  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  length  of  the  curve  between  any  two  points  x  =  a  and 
x  =  b  is 

8=  f(l  +  9a0*.cte=  AC1  +  9&)f-iM1  +  9ot)f- 

PROBLEMS 

1.  Find  the  length  of  y  =  2x  from  #  =  5  to  a  =  10. 

2.  Find  the  length  of  y  =  3  x  +  5  from  a  =  5  to  re  =  20. 

3.  Find  the  length  of  9/  =  lar5  from  (0,  0)  to  (x,  y). 

4.  Find  the  length  of  9  y2  =  4  ar5  from  a;  =  0  to  a;  =  10. 

5.  Find  the  length  of  9  y1  =  4  x*  from  a  =  5  to  X  —  15. 


Art.  19.  —  Area  of  a  Plane  Surface 

Denote  by  A  the  area  of  the  surface  bounded  by  the  continu- 
ous curve  whose  equation  is  y=f(x),  the  lines  x  =  a,  x  —  b, 
and  the  X-axis.     Divide  the  portion  of  the  X-axis  from  x  =  a 

to  x  =  b  into  any  number  of 
equal  parts  Ace.  Construct  rec- 
tangles on  each  Ace  and  the  adja- 
cent ordinates  as  indicated  in 
the  figure.  Denote  by  AA  the 
portion  of  A  included  between 
two  successive  ordinates.  Then 
LA  =  y  •  Ace  -f  $  •  Ace  •  Ay,  where  6 
is  less  than  unity.  This  is  true 
for  all  magnitudes  of  Ace.  When  Ace  becomes  an  infinitesimal, 
Ay  also  becomes  an  infinitesimal.  Hence  the  infinitesimal 
element  of  area  differs  from  the  area  of  the  infinitesimal  rec- 
tangle by  an  infinitesimal  of  a  higher  order,  and  it  follows  that 

x=b  r*h 

A  =  limit  %  y  •  Ax  =  I   y-dx  when  limit  Ace  =  0. 


A  8 


Fig.  9. 


APPLICATIONS 


49 


Example.  —  Find  the  area  of  the  surface  bounded  by  the 

parabola  y2  =  4  x,  the  X-axis,  and  the  lines  x  =  4c,  x  =  9.     For 

this  curve 

Y 


A 


4  AX 

Fig.  10. 


y  =  2  a£  and  ^1  =  2  f  a^  •  cfo  = 
The  area  from  the  vertex  to  the  point  (as,  y)  is 
.4  =  2  fV  .(fo=fa>* 


=  28. 


The  area  from  a;  =  a  to  a?  =  b  is 


J.  =  2  f  ^ .  da?  =  |  (6f  -  a1). 

The  area  of  the  surface  bounded  by  the  parabola  y2  =  4  a;, 
the  F-axis,  and  the  lines  y  =  4,  ?/  =  6  is 

4  =£xdy  =  f£V '  %  -  tV  [V7=  12-67-* 

*  The  areas  of  surfaces  bounded  by  curves  whose  equations  are  not 
known  may  be  found  mechanically  by  means  of  an  instrument  called  the 
planimeter. 


50         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

PROBLEMS 

1.  Find  area  bounded  by  y  =  3x,  the  lines  x  =  0,  x  =  S 
and  the  X-axis. 

2.  Find  area  bounded  by  y  =  5x,  the  lines  x  =  l,  #  =  4 
and  the  X-axis. 

3.  Find   area  bounded   by   y  =  mx  +  n,   the   lines   x  =  a, 
x  =  b  and  the  X-axis. 

4.  Find   area  bounded   by  y  —  mx  -f  n,   the    lines   y  =  a, 
y  =  b  and  the   y-axis. 

5.  Find  area  bounded  by  y2  =  9x,  x  =  0,  #  =  4  and  the 
X-axis. 

6.  Find  area  bounded  by  the  parabola  y2  =  2px,  the  or- 
dinate of  the  point  (x,  y)  of  the  parabola  and  the  X-axis. 

7.  Find  area  bounded  by  the  parabola  y2  =  2px,  the  ab- 
scissa of  the  point  (x,  y)  of  the  parabola  and  the  F-axis. 

8.  Find  area  bounded  by  »*  +  y'2  =  a2  and  the  coordinate 
axes. 

9.  Find  area  bounded  by  y2  =  9  x,  y  =  x  and  x  =  4. 

10.  Find  area  bounded  by  y2  =  9  x,  y  =  2  x  and  y  =  6. 

1 1 .  Find  area  bounded  by  y2  =  4  a  and  ?/  =  ^  ic. 

12.  Find  area  bounded  by  the  parabolas  y2=  4  x  and  x2=Ay. 

Art.  20.  —  Area  of  a  Surface  of  Revolution 

Denote  by  A  the  area  of  the  surface  generated  by  the  revo- 
lution of  the  continuous  curve  y  =  f(x)  from  x  =  a  to  x  =  b 
about  the  X-axis.  A  is  the  limit,  when  limit  Ax  =  0,  of  the 
sum  of    the   areas  of   the    frustums   of  cones   of   revolution 


APPLICATIONS 


51 


generated  by  the  revolution  of  the  parts  of  the  broken  line 
ANB.     Hence  A 


W*+S+Ar)(i+g)**i 


=Hy 


'i+^*-<to 


Fig.  11. 


when  limit  Aa;  =  0. 

Example.  —  Find  the  area 
of  the  surface  generated  by 
the  revolution  of  the  line 
y  —  2x  about  the  X-axis  and 
bounded  by  planes  perpen- 
dicular to  the  X-axis  at  x  =  3 
and  x  —  8. 

Here 

A  =  2tt  C82x(1  +  4)1  •  dx  =  4  V5tt  f^do;  =  i^J  x2 
=  2  V5  tt  •  55  =  110  V5  tt  =  755.592. 

PROBLEMS 

1.  Find  the  area  of  the  surface  generated  by  the  revolution 
of  the  line  y  =  3  x  +  5  about  the  X-axis  and  included  by  the 
planes  perpendicular  to  the  X-axis  at  x  =  0,  x  =  5. 

2.  Find  the  area  of  the  surface  bounded  by  the  revolution 
of  the  line  y  =  c  about  the  X-axis  and  bounded  by  planes 
perpendicular  to  the  X-axis  at  x  =  a,  x  =  b. 

3.  Find  the  area  of  the  surface  of  revolution  generated  by 
the  line  y  =  mx  +  n  revolving  about  the  X-axis  and  included 
by  planes  perpendicular  to  the  X-axis  at  x  =  a,  x  —  b. 

4.  Find  the  area  of  the  surface  generated  by  the  revolution 
of  x%  -f-  2/3  =  a?  about  the  X-axis. 


52 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Art.  21.  —  Volume  of  a  Solid  of  Revolution 

Denote    by   V  the    volume   of  the    solid   bounded   by  the 

surface  generated  by  the 
revolution  of  the  contin- 
uous curve  y  =/(#)  about 
the  X-axis  and  planes  per- 
pendicular to  the  X-axis 
at  x=  a,  x  =  b. 

Denoting  by  AFthe  vol- 
ume of  the  part  of  the 
solid  included  by  planes 
perpendicular  to  the  X-axis 
at  x  and  x  +  Ax, 


Fig.  12. 


and 


Try2  •  Ax  <  A  V<  it  (y  +  A?/)2 .  Ax, 
AV-  try2- Ax  =  6Tr(2y-Ay  +  Ay2)  Ax, 


where  $  is  less  than  unity.  When  Ace  becomes  an  infinitesi- 
mal, Ay  also  becomes  an  infinitesimal.  Hence  in  the  limit 
when  limit  Ax  =  0, 


A  V  =  Try2  Ax,     and     V=  ^-n-y2  Ax 


-■£ 


y2dx. 


Example.  — Find  the  volume  of  the  prolate  spheroid.     The 
prolate   spheroid  is  gener- 
ated  by  the  revolution  of 

the  X-axis.  The  entire 
spheroid  is  comprised  be- 
tween x  =  -f-  a  and  x  =  —  a. 
Hence, 


the  ellipse  —  4-  *-  =  1  about 


APPLICATIONS  53 


/,+°/)2  h2r  ~i+a 


PROBLEMS 

1.  Find  the  volume  of  the  solid  generated  by  the  revolution 

of  the  ellipse  —  -f  ^-  =  1  about  the  Y"-axis.     This  is  called  the 

a-     b2 
oblate  spheroid. 

2.  Find  the  volume  of  the  part  of  the  sphere  x2  -f-  y2  -f  z2  =  25 
between  x  =  2  and  x  =  4. 

3.  Find  the  volume  of  the  solid  bounded  by  the  surface 
generated  by  the  revolution  of  the  parabola  y2  =  2px  about 
the  X-axis  and  the  plane  x  =  a- 

4.  Find  the  volume  of  the  solid  bounded  by  the  surface 
generated  by  the  revolution  of  '—  —  —%  =  1  about  the  X-axis  and 
the  planes  x  —  c,  x  =  d,  where  c  and  d  are  greater  than  a. 

5.  Find  the  volume  of  the  solid  bounded  by  the  surface 
generated  by  the  revolution  of  y  =  3x  +  2  about  the  X-axis 
and  planes  perpendicular  to  the  X-axis  at  x  =  2,  x  =  7. 

6.  Find  the  volume  of  the  solid  bounded  by  the  surface 
generated  by  the  revolution  of  x*  +  y^  =  a2  about  the  X-axis 
and  the  planes  x  =  0,  x  =  a. 


Art.  22.  —  Solids  Generated  by  the  Motion  of  a  Plane 

Figure 

Example.  —  The  ellipsoid  —  -f  -L-\-  —  =  1  may  be  generated 

a2      b2      c2 

by  an  ellipse  whose  center  moves  on  the  X-axis,  whose  plane 


54 


DIFFERENTIAL  AND  INTEGRAL    CALCULUS 


is  perpendicular  to  the  X-axis,  and  whose  axes  in  any  position 

are  the  intersections  of  the 
plane  of  the  generating  ellipse 
with  the  fixed  ellipses, 

a2+b2~   'a^c2'1' 

The   area  of   the   generating 

ellipse    in     any    position     is 

A»  -n-rs-rt.     Since  (x,  rs,  0)  and 

(as,   0,  rt)   are   points   in  the 

L  +  f!-|-^  =  l,  rs  =  -(a2-^and  rt  =  -(a2  -  ar2)*. 
a2      b~     cr  a  a 

Hence,  denoting  the  area  of  the  generating  ellipse  in  terms  of 

x  by  X, 

X=^(a2-x>). 
a2 

Denoting  by  AF  the  volume  of  the  part  of  the  ellipsoid 
included  by  the  generating  ellipses  at  x  and  x  +  Ax, 


Fig.  14. 


9? 


and 


X Ax  >  AV>  (X  -  AX)  Ax, 
AV=XAx-6>AX-Ax, 


be 


where  AX  represents  the  change  of  X  =  ir  —  (a2  —  x2)  corre- 

sponding  to  a  change  of  Ax  in  the  value  of  x  and  6  is  less 
than  unity.  Since  AX  becomes  an  infinitesimal  when  Ax 
becomes  an  infinitesimal,  in  the  limit,  when  limit  Ax  =  0,  the 
volume  of  the  ellipsoid  is 

F=  limit  ~2X.A.c=  C+*X-dx 

x=— a  c/— a 

be  r+a 

=  7T—  I       (a2  —  a2)  dx  =  1 7T  a&c. 

(X  %) -a 


APPLICATIONS 


55 


PROBLEMS 


1.    Find  the  volume  of  the  part  of  the  elliptic  paraboloid 


-  -\ —  =  2  x  included  between  the  planes  x  =  0  and  x  —  5. 


9 


2.    Find  the  volume  of  the  solid  bounded  by  the  hyperboloid 


of  one  sheet 


+K-* 


1  and  the  planes  z  =  3,  z  =  5. 


Fig.  15. 


9   '  4 

3.  Two  equal  semi-circles  x2  =  2  ?-z  —  z2,  y2  =  2  ?-z  —  z2  lie  in 
the    perpendicular    planes    XZ, 

YZ.  The  solid  generated  by  the 
square  whose  center  moves  on 
the  Z-axis,  whose  plane  is  per- 
pendicular to  the  Z-axis,  and 
whose  dimensions  in  any  posi- 
tion are  the  chords  of  intersec- 
tion of  the  plane  of  the  square 
with  the  fixed  semi-circles,  is 
called  a  semi-circular  groin.  Find 
the  volume  of  this  groin. 

Notice  that  the  groin  might  be  defined  as  the  solid  which 
the  two  cylinders  x2  =  2  rz  —  z2  and  y2  =  2  rz  —  z2  have  in 
common. 

4.  Two  semi-circles  x2  =  2  rx z  —  z2,  y2  =  2  r2z  —  z2  of  unequal 
radii  lie  in  the  perpendicular  planes  XZ,  YZ.  Find  the 
volume  of  the  groin  generated  by  the  rectangle  whose  plane  is 
perpendicular  to  the  Z-axis  and  whose  dimensions  are  the 
chords  of  intersection  of  the  plane  of  the  rectangle  with  the 
given  semi-circles.     Take  depth  of  groin  d. 

5.  The  two  parabolas  x2  =  2 pxz,  y2  —  2p2z  lie  in  the  per- 
pendicular planes   XZ,   YZ.     Find  the  volume  of  the  groin 


56 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


generated  by  the  rectangle  whose  plane  is  perpendicular  to 
the  Z-axis  and  whose  dimensions  are 
the  chords  of  intersection  of  the  plane 
of  the  rectangle  with  the  given  parab- 
olas.    Take  depth  of  groin  d. 

6.  Two  circular  cylinders  have  equal 
bases  and  equal  altitudes.  Their  lower 
bases  are  tangent  to  each  other,  their 
upper  bases  coincident.  Find  the  vol- 
ume of  the  solid  common  to  the  two 
fig.  16.  cylinders. 


CHAPTER   V 

SUCCESSIVE  ALGEBKAIO  DIFFEKENTIATION    AND 
INTEGKATION 

Art.  23.  —  The  Second  Derivative 


The  first  derivative  of  /(*),  denoted  by  — f(x)  or  f'(x),  is 

(XX 

in  general  a  function  of  x.     The  first  derivative  of  the  first 

d  r  d         ~\  d2 

derivative,  —  — fix)   ,  is  denoted  by  — 0f(x)  or /"(#),  and  is 
dx[_dx        J  dx2 

called  the  second  derivative  of  f(x).     If  f(x)  is  denoted  by  y, 

the  first  and  second  derivatives  are  written  -2  and  J- 

dx  dx* 

For  example, 

if  f(x)  =  xi-7x  +  7,  f'(x)  =  3x2-7,  f"(x)  =  6x. 

Geometrically  ^  measures  the  slope  of  the  tangent  to  the 

curve  whose  eoit  is  „  =/(*)  at  (*,,).    Hence  £»  -  g 
.  dx\dxj      dxr 

measures  the  rate  01  change  Y 

of  the  slope  of  the  tangent. 

When    — |   is  positive,   -^ 
dx2  dx 

or  the  slope  of  the  tangent 

increases    as    x    increases. 

From  the  figure  this  is  seen 

to  be  the  case  for  the  part 

cde  of  the  curve,  that  is, 

when  the  curve  is  concave 


Fig.  17. 


57 


58  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

upward.    When  ^  is  negative,  the  slope  of  the  tangent  ele- 
cta2 
creases  as  x  increases.     This  is  the  case  for  the  part  abc  of  the 

curve;   that   is,  when   the  curve  is  convex  upward.      When 
— ^  =  0,  the  tangent  is  stationary.     A  point  of  the  curve  where 

(XX> 

the  tangent  is  stationary  is  called  a  point  of  inflection,  since 
at  such  a  point  the  curve  changes  its  direction  of  curvature. 

If  the  relation  between  the  distance  passed  over  and  the 
time  of  the  motion  of  a  body  is  expressed  by  the  equation 
s  =  fit),  —  is  the  velocity  at  any  time  t.     If  the  velocity  has 

(XL 

different  values  for  different  values  of  t,  represent  the  velocity 

at  time  t0  by  v0,  at  time  t  by  v.     The  average  rate  of  change  of 

velocity  in  the  interval  of  time  t  —  t0  is      —   °»     Calling  the 

*  ~~  ^°  Av 

interval  of  time  At,  the  change  in  velocity  Av.  the  ratio  — 

At 
measures  the  average  rate  of  change  of  velocity  in  the  interval 

of  time  At     This  is  true  whatever  may  be  the  magnitude  of 

the  interval  of  time  At.     The  limit  of  this  ratio  when  limit 

At  =  0  is  the  actual  rate  of  change  of  velocity  at  time  t,  for 

the   interval   of    time   during   which   the  rate  of   change  of 

velocity  might  vary  continually  decreases.     The  actual  rate  of 

change  of  velocity  is  called  acceleration.     Hence  acceleration 

,.    .,    Av     dv      d  ds     d2s       ^  ,       .„  h~^o,9 

=  limit  —  =  —  =  —  —  =  —.      For   example,   if  s  =  16.08 12, 
At      dt      dt  dt      dt2  ^    ' 

denoting  the  acceleration  by  a,  a  =  32.16. 
As  a  direct  consequence  of  the  definition 

b^-sGb'w}  lt  foIlows  ^/J>>=I'<'>+<* 

For  example,  let  it  be  required  to  find  the  function  whose 
second  derivative  is  2x  —  o.  Denoting  the  function  by  fix), 
-fr2f(x)  =  2x  +  5.    Integrating,  -£-/(«)  =  x2-5x+C1.     Inte- 

CiXi  CiX 


SUCCESSIVE  DIFFERENTIATION  59 

grating  again,  f(x)  —  \t?  —  ^x2  -\-  Cxx  -\-  C2-  C\  and  C2  are  two 
arbitrary  constants  of  integration,  whose  values  become  known 
if  the  function  and  its  first  derivative  are  required  to  take 
given  values  for  given  values  of  x.  For  instance,  if  it  is  re- 
quired that  fix)  —  10  when  x  =  2  and  — f(x)  =  1  when  x  =  0, 

(XX 

Cx  —  \.  and  C2  =  15  J.     Under  these  conditions,  if 

£j(x)  =  2x-  5,  /(*)  =  i^-f^-f  x  +  151. 

PROBLEMS 

Find  the  first  and  second  derivatives  of, 

1.    3X2.  2.    4ar>-5.  3.   2x3-7x. 

1                     k        1  a 

4.    — •  5. •  6. 


X  1  —  X  1  +  X2 

Find  between  what  values  of  x  the  curves  which  represent 
the  following  equations  are  convex  upward,  concave  upward, 
and  find  the  points  of  inflection, 

7.  2,  =  ^ -7* +  7.    s.  y=~7     9-  y=f-^+2- 

Determine  the  first  and  second  derivatives  of  y  with  respect 
to  x  in  the  implicit  functions, 

10.   x2  +  y2  =  r2.     The  first  derivative  is  -»  = 

dx         y 

Differentiating  both  sides  of  this  equation  with  respect  to  x, 


d2y                   da; 

dx2  ~~           ?/2 

Substituting  ^  = 
dx 

x                             d  u 

,  there  results  — %  = 

y                          dxr 

r2 

a2     62 

12.    %-t      1. 

a2     bs 

13. 

13.    y2  =  2px. 


60  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Find  ^1  from  the  following  equations, 
dx2 

!4.    fM\  =1/3.     Differentiating  both  sides  of  this  equation 
with  respect  to  x,  2  j|  g  =  3  y* J|  whence  g  =  |  y\ 

15.    ^  =  2/2.  16.    ft.*  17.    f  =  l 

dx  dx  dx     x 

"■  (%}->■        '•■  (I)"-" 

20.  The  equation  f(x,  y)  =  0  defines  ?/  as  a  continuous  func- 
tion of  x.     Consequently   it   also   defines  a;  as  a  continuous 

function  of  y.     It  has  been  shown  that  -/  =  —-•     Prove   that 

9  dx     dx 

^  dy 

d2y  _       dy2 

dx2~      fdx^z 

Afj 

A  body  moves  in  such  a  manner  that  the  relation  between 

s  and  t  is  expressed  by  the  following  equations.     Find  velocity 
and  acceleration  at  time  t.     a  and  u  are  constants. 

21.  s  =  u-t  +  ±a-t2.  23.    s  =  u-t2  +  a>t\ 

22.  s  =  u-t  —  \a>t2.  24.    s  =  u-t2-a-f. 

Determine  the  functions  whose  second  derivatives  are, 

25.  3 a; +  2,   knowing    that   when  8 si,  f(x)  =  10;   when 

26.  ox  —  lx1,  knowing  that  when 

*  =  0, /(»)  =  <>,  £/(■)- 1. 

27.  Find  the  curve  through  the  origin  and  making  an  angle 

of  45°  with  the  X-axis  at  the  origin,  knowing  that  ~^=2x  +  5. 

dx2 


SUCCESSIVE  DIFFERENTIATION  61 

28.  Find  the  curve  through  (5, 7)  and  parallel  to  the  X-axis 

d2v 
at  x  =  3,  knowing  that  —2  =  2  x2  —  4  x. 
dx2 

29.  A  body  starts   moving  with   a  velocity  u  and  has   a 
uniform  acceleration  a.     Find  the  relation  between  s  and  t. 

The  conditions  of  the  problem  are  expressed  by  —  —  a,  and 

ds  ^ 

when  t  —  0,  s  =  0,  —  —u.     By  integration 

^  =  a't+  Ci,  s  =  -ia.*2  +  Ci-*  +  <72. 

Since  s  =  0  and  —  =  u  when  t  =  0,  Ci  =  w  and  C2  =  0.    Hence 
dt 

the  result  is  s  =  ut  +  \  at2. 

30.  A  body  starts   moving  with  a  velocity  u  and   has  a 
uniform  acceleration  —  a.     Find  the  relation  between  s  and  £. 

31.  A  horizontal  beam  has  one  end  fixed.     The  bending  of 
the  beam  due  to  a  weight  P  causes  an  elongation  of  the  upper 
surface  of  the  beam  and  a  com- 
pression of  the  lower  surface.  j^ 

Some  intermediate  surface   re-     J^_Z___ZZZZ — 
mains  unchanged  in  length  and      J-  ,  i  j 

is  called  the   neutral   surface.  l_i_* 

The    figure    shows    a  vertical  f?% 

longitudinal     section     of     the  ^^ 

Fig.  18. 

upper,  lower,  and  neutral  sur- 
faces of  the   beam.    This   section  of  the  neutral   surface   is 
called  the  elastic  curve.     In  text-books  on  Strength  of  Mate- 
rials it  is  proved  that  for  any  point  (x,  y)  in  the  elastic  curve, 

d2v 
EI  -—  =  —  P-  x,  where  E  is  a  constant  depending  on  the  mate- 
dx2 

rial  of  the  beam,  I  a  constant  depending  on  the  cross-section 
of  the  beam.     By  the  nature  of  the  problem  y  =  0  when  x  —  0, 


62  DIFFERENTIAL   AND  INTEGRAL   CALCULUS 

and.  ty  =  0  when  x  =  Z.     Determine  the  equation  of  the  elastic 

dx 
curve  and  the  maximum  deflection  of  the  beam. 

32.  Suppose  the  load  uniformly  distributed  over  the  beam 
of  Problem  31.    If  the  load  per  linear  unit  is  w  and  the  origin 

,    -,  of  coordinates  is  taken  at  the 

i"T~j~  free    end   of    the   beam,   it    is 

't^'IQOQQQQq^q-^  proved  in  Strength  of  Materials 

iii — ■ — "^^^rc^Qcx   ^a^  ^or  any  p°^n^  (x>  y)  m  *ne 

i      i                           2/t^i^7                                     d2v      —  w  •  x2 
i  ^1      elastic   curve  EI—^-  = 

fig.  19.  By  the  nature  of  the  problem 

-^  =  0  when  x  =  I,  and  y  =  A  when  x  =  I,  where  A  represents 

dx 

the  deflection  of  the  free  end.     Determine  the  equation  of  the 

elastic  curve. 

33.  A  horizontal  beam  rests  on  two  supports  the  distance 
between  which  is  I.  If  a  load  P  is  placed  at  the  middle  of  the 
beam  and  the  origin  of  coordinates  is  taken  at  the  left  support, 
it  is  proved  in  Strength  of  Materials  that  for  any  point  (x,  y) 

d?v      Px 
in  the  elastic  curve  j£7^  =  — -.     By  the  nature  of  the  prob- 

dr       2  ' 

lem  -^  =  0  when  x  =  Xl,  and  y  =  0  when  x—0.     Find  the 
dx 

equation  of  the  elastic  curve  and  the  maximum  deflection. 

34.  The  beam  of  Problem  33  supports  a  uniform  load.  If 
w  is  the  load  per  linear  unit,  it  is  proved  in  strength  of  mate- 
rials that  for  all  points  in  the  elastic  curve 

jPT^2y  —  ?£l®  —  wx2 
dx^~~Y       2  ' 

By  the  nature  of  the  problem  -^  =  0  when   x  =  \l,   and  y  =  0 

dx 


S  UCCESSI VE  DIFFER  EN  TIA  TION 


63 


when  x  =  0.     Find  the  equation  of  the  elastic  curve  and  the 
maximum  deflection. 

35.  A  circular  disc  whose  weight  is  W  is  free  to  turn  about 
a  horizontal  axis  through  the  center  of  the  disc  and  perpen- 
dicular to  the  plane  of  the  disc.  A  cord  wound  around  the 
circumference  of  the  disc  has  a  weight 
P  attached  to  its  free  end.  In  Mechan- 
ics it  is  proved  that 

dM  =        P-r-g 

dt2     W>k2  +  P-r* 

where  g  is  the  earth's  gravity  constant, 
r  is  the  radius  of  the  disc  in  feet,  k  is 
a  constant  depending  on  the  size  of  the 
disc,  and  0  is  the  angle  in  radians 
through  which  the  disc  turns  in  t  seconds, 
between  6  and  t. 

In  the  following  problems  find  -f- 

dx 

d2v 

36.  —4=  y*.     Multiply  both  sides  of  the  given  equation  by 
dx2 

®  and  integrate.     There  results 
dx 


Fig.  20. 

Find  the   relation 


dy  i  d2y 
dx    dx2 


dy 


dx 


tf.?2.     and    ifan-J^+a 


#Y_  1  ,A 


dxj 


37.    ^  =  2^  +  ^. 
dx2        dx 


38. 


dx2 


2y2-5y. 


Art.  24.  —  Maxima  and  Minima 


A  continuous  function  f(x)  increases  as  x  increases  when  its 
first  derivative  — f(x)  is  positive,  and  decreases  as  x  increases 
when  its  first  derivative  is  negative.     If  f(x)  changes  from  an 


64         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

increasing  to  a  decreasing  function  when  x  =  x0,  the  value  of 
fix)  is  greater  when  x  =  x0  than  it  is  just  before  x  reaches  x0, 
and  also  greater  than  it  is  just  after  x  passes  x0.  This  is 
expressed  by  saying  that  f(x)  has  a  maximum  value  when 
x  =  x0.     At   a  maximum   value   of  f(x)   therefore   the   first 

derivative  — fix)  changes   sign  from  +  to  —  for  increasing 
dx 

values  of  x. 

If  fix)  changes  from  a  decreasing  to  an  increasing  function 
when  x  =  x0,  the  value  of  fix)  is  less  when  x  =  x0  than  it  is 
just  before  x  reaches  xQ  and  also  less  than  it  is  just  after  x 
passes  Xq.  This  is  expressed  by  saying  that  fix)  has  a  mini- 
mum value  when  x  =  xQ.  If  fix)  changes  from  a  decreasing 
to  an  increasing  function,  the  first  derivative  must  change  sign 
from  —  to  -K     Hence  at  a  minimum  value  of  fix)  the  first 

derivative  — fix)  changes  sign  from  —  to  +  for  increasing 
dx 

values  of  x. 

An  algebraic  function  can  change  sign  only  by  passing 
through  zero  or  by  passing  through  infinity.    Hence  when/(#) 

has  a  maximum  or  a  minimum  value,  —  f(x)=Q  or  — f(x)  =  ao. 
d  dx  dx      ' 

If   —f(x)  =  0   and  fix)   has   a  maximum   value,    — f(x) 
dx  dx 

changes  sign  from  -f  to  —  by  passing  through  zero.     Hence 

—  f(x)  is  a  decreasing  function,  and  its  first  derivative,  which 
dx  2 

is  the   second  derivative   of  f(x),  that  is  — -^f(x),  must  be 

negative.  d7? 

If    -j-f(x)  =  0   and  f(x)   has   a  minimum   value,    —fix) 
dx  dx 

changes  sign  from  —  to  +  by  passing  through  zero;   hence 

— f(x)  is  an  increasing  function,  and  its  first  derivative,  which 
dx 

is  the  second  derivative  off(x),  must  be  positive. 

These  results,  obtained  analytically,  may  also  be  directly 


8  UCCESSI VE  DIFFERENTIA  TION 


65 


obtained  from  the  graph  of  y  =  f(x).  For  increasing  values  of 
x  the  point  generating  the  continuous  curve  which  represents 
the  equation  y  =f(x)  can  cease  moving  away  from  the  X-axis 
and  start  moving  towards  the  X-axis  only  in  one  of  two  ways, 
either  by  tending  to  move  parallel  to  the  X-axis,  as  at  Plf 


Fig.  21. 


which  requires  that  -^  =  0,  or  by  tending  to  move  parallel  to 

(XX  -j 

the  Y-axis,  as  at  P2,  which  requires  that  -^=oo.  At  P3, 
j  dx 

while  —  =  0,  y  is  neither  a  maximum  nor  a  minimum.  At  P4, 
dx 

while  -&  =  oo,  v  is  neither  a  maximum  nor  a  minimum.  An 
dx         '  * 

examination  of  the  figure  shows  that  in  all  cases  for  increasing 

values  of  x  the  slope  of  the  tangent,  that  is  — ,  changes  sign 

dx 

from  -f  to  —  at   a  maximum,  from  —  to  -f-  at  a  minimum. 

At  a  maximum,  when  -j-  =  0,  the  curve  is  convex  upward,  and 

hence  — |  is  negative ;  at  a  minimum,  when  -^  =  0,  the  curve 

is  concave  upward,  and  -^  is  positive.     From  the  figure  it  is 
(xx~ 

evident  that  maximum  and  minimum  values  must  occur  alter- 
nately and  that  a  minimum  may  be  greater  than  a  maximum. 
From  this  investigation  is  inferred  the  following  method  of 
examining  a  function  of  one  variable  fix)  for  maximum  and 


66  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

minimum  values.     Find  the  roots  of  the  equations  —  f(x)  =  0 

d 
and  — f(x)  =  ao.      If,   for   increasing   values   of  x,  —  /(as) 

dx  dx 

changes  sign  from  -f-  to  —  when  x  passes  one  of  these  roots, 

this  root  makes  fix)  a  maximum,    if  —f(x)   changes   sign 

dx 

from  —  to  -f-  when  x  passes  one  of  these  roots,  this  root  makes 

fix)  a  minimum  ;    if  —  fix)  does  not  change  sign  when  x 
dx 

passes  through  one  of  these  roots,  this  root  makes  fix)  neither 
a  maximum  nor  a  minimum. 

The  roots  of  the  equation      ,    -  =  0  which  make     *^  '  nega- 

dx  dx2 

tive  make  f(x)  a  maximum ;    the  roots  of   -J-LJ.  =  o  which 

d2f (x)  • 

make    J  K  '  positive  make  fix)  a  minimum. 
dx2 

Example    I.  —  Examine    x*  —  7  x  -f  7    for    maximum    and 

minimum  values.      Here  f(x)  =  x3  —  7  x  -j-  7,  '  =  3  x2  —  7, 

d2f(x) 
•  j\ '  =  6 x.     Equating   the   first   derivative  to  zero,  Sx2  =  7 

—  d2fix) 

and   x  =  ±  V|.      x  ==  -f  Vj   makes       ,*    positive   and  f(x) 

.-  d2fix) 

=  —  .145,  a  minimum.      #  =  —  V -J  makes     \  ,    negative  and 

/(#)  =  14.145,  a  maximum. 

/«.  _i_  o\3 
Example    II.  —  Examine    y  = v     '     '    for    maxima    and 

(x  -  3)- 

minima.       Here  ^  =  (a?+/2)2(al~ 13).       The    first  derivative 
dx  (x  —  3)3 

equated  to  zero  gives  x  =  —  2,  x  =  -f  13 ;  the  first  derivative 
equated  to  infinity  gives  x  =  +  3.  When  a;  is  just  less  than 
—  2,  the  signs  of  the  three  factors  of  -M.  are  "*  and  _^  is 
positive;  when  a  is  just  greater  than   —2,  the  signs  of  the 

factors   of   ^   are   itn  and  ^   is   still   positive.     Since  ^ 
dx  —  dx  dx 


SUCCESSIVE  DIFFERENTIATION  67 

does  not  change  sign  when  x  passes  through  —  2,  y  is  neither 
a  maximum  nor  a  minimum  when  x  =  —  2. 

When  a;  is  just  less  than  +13,  the  signs  of  the  factors  of 

_Z  are  and  - ?  is  negative;  when  a;  is  just  greater  than 

do?  +  da;  ,  _  , 

+ 13,  the  signs  of  the  factors  of  JL  are    *"      and  -M.  is  posi- 
,  da;  +  dx 

tive.     Since  -^  changes  sign  from   —   to   +    when  x  passes 

(XX 

through  13,  y  is  a  minimum  when  x  =  13.     This  minimum  is 

2/  =  33|. 

When  x  is  just  less  than  -j-  3,  the  signs  of  -~  are  and 

-3t  is  positive ;    when  x  is  just  greater  than  +  3,  the  signs  of 
da; 

~  are and  -~  is  negative.      Since  -~  changes  sign  from 

dx  -f-  dx  dx 

+  to  —  when  x  passes  through  +3,  y  is  a  maximum  when 
x  =  -f-  3.     This  maximum  is  y  =  oo. 


PROBLEMS 

Examine  the  following  functions  for  maximum  and  mini- 
mum values : 


1. 

a*  — &»  +  & 

g 

x-1 

2. 

ar3  —4*4-7. 

(x  +  2f 

3. 

2a?-5x. 

10. 

(a;-l)4(a;  +  2)3. 

4. 

xi-2x\ 

11. 

y  =  2  X  —  X2. 

5. 

ix*-§tf-i 

x2  +  2x. 

12. 

y  =  2  Rx  —  x2. 

6. 

tf-5x2-10x  +  4:. 

13. 

y  =  ^2(2ax-x2). 

7. 

(x-1)2 
(x  +  lf 

X 

14. 
IS. 

CI 

y  —  a?  +  3  x  —  5*. 

3a;-5 

?/  =3 

1  +  ar*  '      (2a;-3)a 


68  DIFFERENTIAL  AND   INTEGRAL   CALCULUS 


16.  y  =  10  V8  x  —  x\  Suggestion.  10  V8  x  —  x2  is  a  maxi- 
mum when  8  a  —  x2  is  a  maximum,  a  minimum  when  Sx  —  x2 
is  a  minimum.  Hence  constant  factors  may  be  dropped  and 
the  radical  sign  removed  before  forming  the  first  derivative. 


17.    -VStf-lOx  +  G.       18.    J- — !•      19.    3vW-10a>». 

*  x  -f-  o 

20.  Show  that  x3  —  3  x2  +  6  x  has  neither  a  maximum  nor  a 
minimum  value. 

21.  Show  that  9^SZ —  is  a  maximum  value  of  ax2  +  2  bx  +  c 

a 
when  a  is  negative,  a  minimum  value  when  a  is  positive. 

22.  Divide  a  into  two  parts  whose  sum  is  a  minimum. 

23.  Divide  a  into  two  parts  such  that  the  sum  of  their 
squares  is  a  minimum. 

24.  Divide  a  into  two  parts  such  that  their  product  is  a 
maximum. 

25.  Divide  a  into  two  parts  such  that  the  sum  of  their 
square  roots  is  a  maximum. 

26.  From  a  square  sheet  of  tin  18  inches  on  a  side  equal 
squares  are  cut  at  the  four  corners.  From  the  remainder  of 
the  sheet  of  tin  a  vessel  with  open  top  is  formed  by  bending 
up  the  sides.  Find  side  of  small  squares  when  the  vessel 
holds  the  greatest  quantity  of  water. 

27.  From  a  rectangular  sheet  of  tin  3  feet  by  2  feet  equal 
squares  are  cut  at  the  four  corners  and  a  box  with  open  top 
formed  by  turning  up  the  sides.  Find  sides  of  squares  cut 
off  when  contents  of  box  are  greatest. 

28.  A  box,  square  base  and  open  top,  is  to  be  constructed 
to  contain  108  cubic  inches.  What  must  be  its  dimensions  to 
require  the  least  material  ? 


SUCCESSIVE  DIFFERENTIATION  69 

29.  A  circular  cylindric  standpipe  is  to  be  built  to  hold 
10,000  cubic  feet  of  water.  Find  altitude  and  diameter  of 
base  which  require  least  material. 

30.  Find   the   shortest    distance   from    (3,   5)   to   the   line 

?  _  V  =  1 
2     3 

31.  Find  the  shortest  distance  from  (7,  8)  to  the  parabola 
y2  =  4  x. 

32.  Find  the  shortest  line  that  can  be  drawn  through  (a,  b) 
meeting  the  rectangular  axes. 

33.  A  Norman  window  is  composed  of  a  rectangle  sur- 
mounted by  a  semicircle.  Find  the  dimensions  of  the  window 
so  that  with  a  given  perimeter  the  window  admits  the  greatest 
amount  of  light. 

34.  Two  trains  are  running,  the  one  due  east  at  30  miles 
per  hour,  the  other  due  north  at  40  miles  per  hour.  When 
the  first  train  is  30  miles  from  the  intersection  of  the  tracks 
the  second  is  20  miles  from  this  point.  Find  the  least  dis- 
tance between  the  trains. 

35.  A  person  in  a  boat  3  miles  from  the  nearest  point  of 
the  beach  wishes  to  reach  in  the  shortest  time  a  place  5  miles 
from  that  point  along  the  beach.  If  he  can  walk  5  miles  an 
hour,  but  row  only  4  miles  an  hour,  where  must  he  land  ? 

36.  The  strength  of  a  rectangular 
beam  varies  as  the  product  of  the 
breadth  and  the  square  of  the  depth. 
What  are  the  dimensions  of  the 
strongest  beam  that  can  be  cut  from 
a  log  whose  cross-section  is  a  circle 
18  inches  in  diameter?  Strength  pre- 
vents breaking.  Fig.  22. 


70  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

37.  The  stiffness  of  a  rectangular  beam  varies  as  the  product 
of  the  breadth  and  the  cube  of  the  depth.  Find  the  dimen- 
sions of  the  stiffest  beam  that  can  be  cut  from  a  log  whose 
cross-section  is  a  circle  18  inches  in  diameter.  Stiffness  pre- 
vents bending. 

38.  The  bending  moment  of  a  simple  beam  whose  length  is  I 

when   the   uniform   load    per 
;  linear  unit  is  w  is 


Sl'  M  =b  \  wlx  —  \  wx2, 

~r      at  the   point  whose   distance 
from   the   left   support   is    x. 
Find  where  the  bending  moment  is  a  maximum. 

39.  The  distance  between  two  lights  A  and  B  is  d.  The 
intensity  of  A  at  unit's  distance  is  a,  of  B  is  b.  If  the  inten- 
sity of  a  light  varies  inversely  as  the  square  of  the  distance, 
find  the  points  between  the  lights  of  maximum  and  minimum 
illumination. 

40.  If  the  illumination  varies  as  the  sine  of  the  angle  under 
which  light  strikes  the  illuminated  surface  divided  by  the 
square  of  the  distance  from  the  source  of  light,  find  the  height 
of  an  electric  light  directly  over  the  center  of  a  circle  of  radius 
r  when  the  illumination  of  the  circumference  is  greatest. 

t2 

41.  If  c2r-f--  is  the  total  waste  per  mile  going  on  in  an 

r 

electric  conductor,  r  resistance  in  ohms  per  mile  of  conductor, 
c  the  current  in  amperes,  t  a  constant  depending  on  interest  on 
investment  and  depreciation  of  plant,  find  the  relation  between 
resistance  and  current  when  the  waste  is  a  minimum. 

42.  If  v  is  the  velocity  of  an  ocean  current  in  knots  per 
hour,  x  the  velocity  of  a  ship  in  still  water,  and  if  the  quantity 
of  fuel  burnt  per  hour  is  proportional  to  x3,  find  the  value  of  x 


SUCCESSIVE  DIFFERENTIATION 


71 


which  makes  the  consumption  of  fuel  a  minimum  for  a  run  of 
s  miles. 

43.  Given  n  voltaic  cells  of  E.  M.  F.  e  and  internal  resist- 
ance r,  to  find  the  way  in  which  the  cells  should  be  arranged 
to  send  a  maximum  current  through  a  given  external  resist- 
ance R. 

Let  x  cells  be  placed  in  series,  then  the  current  I 


xe 


x2r 


+  n 


44.   The  equation  of  the  path  of  a  projectile  is 
y  =  x  tan  6 


2w2cos26>' 

where  0  is  the  angle  of  projection,  u  the  velocity  of  projection, 
g  =  32.16.    The  range,  the  value 


of  x  when   y  =  0,  is 


n'2sm(20) 
9 


Fig.  24. 


a.  Find  the  greatest  height. 

b.  Find  the  angle  of  projec- 
tion which  gives  the  greatest 
height  for  a  given  velocity. 

c.  Find  the  angle  of  projection  which  gives  the  greatest 
range  for  a  given  velocity. 

45.   Find  the  dimensions  of  the  isosceles  triangle  of  maxi- 
mum area  that  can  be  inscribed  in  a  circle  of  radius  r. 

Calling  altitude  of  triangle   x,  base  2  y,  y2  =  2rx  —  x2)  and 
area  A  =  x  V2  rx  —  x2.     A  is  a  maximum 
when  A'  =  2  rx3  —  x4  is  a  maximum. 


dA' 
dx 


=  6rx?  —Ax3, 


which  is  zero  when  x  =  f  r.  The  first 
derivative  changes  sign  from  +  to  — 
when  x  passes   through  j-r.     Therefore 


72 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


the  area  of  the  inscribed  triangle  is  a  maximum  when  its  alti- 
tude is  §  r. 

46.    Find  the  dimensions  of  the  rectangle  of  maximum  area 

x2      v2 
that  can  be  inscribed  in  the  ellipse  — \-<L  —  l.     The  area  of 

or     ¥ 

the  rectangle  is  A  =  4xy, 
which  is  a  maximum  when 
A'  =  xy  is  a  maximum. 

dA'        dy  , 

=  flj-i  -f-  y. 

dx         dx 


Fig.  26. 


From  the  equation  of  the 
4'— $*    By  sub- 


ellipse 


dx 


ay 


dA'     d2v2  —  tfx2 

stitution, =  — - — - — '—.     Hence   A'  is   a   maximum   when 

dx  ay 


a2y2  _  yLrg  _  o      Combining  this  equation  with  the  equation  of 

the  ellipse,  x  =  -^  y  =  — ~,  and  the  area  of  the   maximum 

V2  V2 

rectangle  is  2ab. 

47.   Find  the  area  of  the  maximum  rectangle  that  can  be 
inscribed  in  the  parabola  y2  =  2px  whose  limiting  coordinates 

are  a,  b. 

48.  Find  the  dimensions  of 
the  cylinder  of  revolution  of 
maximum  volume  that  can  be  in- 
scribed in  a  sphere  of  radius  r. 

Calling  the  radius  of  the  base 

of  the  cylinder  x,  the  altitude 

^  V>  (x,  V)  is  a  point  of  the  circle 

x*  +  V2  =  9s  which  generates  the 

Fig.  27.  sphere.     Calling  the  volume  of 


SUCCESSIVE  DIFFERENTIATION  73 

the  cylinder  V,   V  —  2  nyx2  =  2  try  (r2  —  y2).     V  is  a  maximum 

d  V  t 

when  —  =  2  ttt2  —  6-n-y2  =  0,  that  is  when  y  =  — ,  since  this 

dy  d>V  ^ 

value  of  y  makes  — -  =  —  12  -n-y  negative.     The  maximum  is 

4  dy 

V= z.irr3.      The   ratio   of  the   volume   of   the   maximum 

3V3  1 

inscribed  cylinder  to  the  volume  of  the  sphere  is 

V3 

49.  Find  the  cone  of  maximum  volume  that  can  be  inscribed 

in  a  sphere  of  radius  r. 

50.  Find  the  cylinder  of  maximum  volume  that  can  be 
inscribed  in  a  cone,  altitude  h,  radius  of  base  r. 

51.  Find  the  maximum  cylinder  that  can  be  inscribed  in 
the  prolate  spheroid. 

52.  Find  the  maximum  cylinder  that  can  be  inscribed  in 
the  oblate  spheroid. 

53.  Find  the  cone  of  maximum  volume  that  can  be  in- 
scribed in  a  paraboloid  of  revolution,  the  limiting  point  of  the 
generating  parabola  y2  =  2 px  being  (a,  b),  the  vertex  of  the 
cone  at  intersection  of  axis  of  paraboloid  with  base  of 
paraboloid. 

54.  Of  all  right  circular  cones  whose  convex  surface  is  the 
same  find  the  dimensions  of  that  whose  volume  is  greatest. 

dV     ,    f0n,  dy 


V=\ 7ry2x,  —-  =  liT[2xy^-  +  x2)  and  vyVx2  +  yl  =  constant. 
dx  \       dx        J 

arV  +  v4  =  constant,     and     -^  =    t—-%    •  /j\ 

dx      xr  +  2  yl  /  JL  \ 

Hence  ^^\J2-tlzft\.     Equating  the  /    \\ 

dx            \  x2  +  2y2  J  I              \ 

first    derivative    to    zero,    x  =  y  V2,   which  /    ^_j___^  \ 

makes  V  a  maximum  since  —  changes  sign  C           ^--y — 3 

dx  \^ ^ 

from  -f  to  —  when  x  passes  through  y  V2.  Fig.  26. 


74  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

55.  Of  all  right  circular  cylinders  whose  convex  surface 
plus  the  surface  of  the  lower  base  is  constant,  find  the  dimen- 
sions of  that  whose  volume  is  greatest. 

56.  The  distance  between  the  centers  of  two  spheres  of 
radii  r  and  R  is  d.  Find  on  the  line  joining  the  centers  the 
point  from  which  the  greatest  amount  of  spherical  surface  is 
visible. 

57.  Find  the  axis  of  the  parabola  of  maximum  area  that 
can  be  cut  from  a  right  circular  cone,  radius  of  base  r, 
altitude  h. 

Art.  25.  —  Derivatives  of  Higher  Orders 

It  has  been  found  convenient  to  call  the  first  derivative  of 

the  first  derivative  of  f(x)  the  second  derivative  of  fix),  and 

d2 
to  denote  it  by  — -2f(x)  or  f"(x).     In  like  manner  it  is  found 

convenient  to  call  the  first  derivative  of  the  second  derivative 

the  third  derivative  of  fix),  and  to  denote  it  by  the  symbol 

d3 

-—/(»)  or  /'"(#).     By  an   extension   of  the   same   notation 

&Xi 

fiv(x)>  fv(®)>  —f  /"(*)  denote  the  fourth,  fifth,  ••,  nth  deriva- 
tives of  f(x). 

For  example,  if  fix)    =  x4  +  3  X*  -  7  x2  -  27  x  -  18, 

/'(*)    =±x3  +  9xi-Ux-27, 
fix)  =12z2  +  18z-14, 
/'"(*)  =  24*  +  18, 
f»(x)  =  24, 
JT(*)  =0. 

It  follows  immediately  from  the  definition  that  the  integral 
of  any  derivative  is  the  next  lower  derivative. 


SUCCESSIVE  DIFFERENTIATION  75 

For  example,  if 

fly(x)  =  24:X-lS, 
f"'(x)  =  12xi-18x  +  Cl, 
f"(x)  =4rXs-9x2+C1-x+C2, 

f(x)    =  i«"  -  fa4  +  i  Q  •  a8  +  J  02  •  a?  +  C3  •  x  +  C4. 

The  arbitrary  constants  (7W  <72,  <73,  C4  become  known  if  the 
function  f(x)  and  its  first  three  derivatives  f'(x),  /"(#),  f'"(x) 
are  required  to  take  given  values  for  given  values  of  x. 

The  successive  derivatives  f'(x),  f"(x),  f'"(x),  •  ••,  are  called 
derivatives  of  the  first,  second,  third,  •  •  •,  orders.  If  fn(x)  is 
given,  f(x)  is  found  by  n  successive  integrations,  and  the 
general  expression  for  f(x)  must  therefore  contain  n  arbitrary 
constants.* 

PROBLEMS 

Form  the  derivatives  of  the  first  three  orders  of, 

1.   ^_6a2  +  15.  2.    3x2  +  4z-7.  3.   — !— • 

1  —  a; 

Find  f(x)  when, 

4.  f'"(x)  =x2-5x.  6.  f"(x)  =  16 a?  +  7. 

5.  /IV («)  =  !  +  x.  7.  /"(»)  =  10. 

Art.  26.  —  Evaluation  of  the  Indeterminate  Form  - 

The  ratio  -  may  have  any  value  whatever;  that  is,  the 
value  of  the  ratio  is  indeterminate.     If,  however,  the  ratio  of 

*In  Lagrange's  notation  the  successive  derivatives  of  f(x)  are  denoted 

by  /'(»),  /"(*)i  /'"(*)•"'>    in  Cauchy's  notation,  by  Df(x),  D2f(x), 

Dzf(x)  •  •  • .     Newton  denotes  — -  by  s. 

cit 


76  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

two  functions  "^  takes  the  form  -  for  some  special  value 

#M  °        /(a?) 

of  x,  such  as  a;  as  a,  the  true  value  of   tjtt  when  a;  =  a  is 

denned  as  the  limit  of  the  ratio  /(a  +  Ax)  when  limit  Ax  =  0. 

<£(a  -f-  Ax) 

Since  by  hypothesis  /(a)  =  0  and  <f>  (a)  =  0, 

/(q  +  AaQ-/(tt) 

limit  J)    ^ {  =  limit  —, — — — r TT^^Trri' 

<f>(a  -f  Ax)  <fr(tt  +  Aff)-<fr(g)     <£'(a) 

Ax 

when  limit  Ax  =  0.     Hence  by  the  definition  the  true  value  of 

£+-t  when  x  =  a  is  ^-^,  where  /'(a)  and  fi(a)  denote  the 
<f>(x)  <j>'(a) 

values  of  f'(x)  and  <j>'(x)  when  a  is  substituted  for  x. 

If  |TW  _     the  same  analysis  shows  that  the  true  value  of 
<£'(<x)      0 

/(a)  =/M  -£M      in  general,  the  true  value  of  ^  is  the 

0(a)     *'(«)     *"(«)       *A  ,      .  *(a) 

first  determinate  ratio  ^  ^  '    of  derivatives  of  the  same  order. 

r(a) 

Ex.  — Find  the  true  value  of     f~?X  +  2    'when  a?  =  fc 

or  —  ar  —  x  +  1 

-„.       /(»)  ar>-3a;  +  2        0,  , 

Here  ^-^    =  -- — -^— -  =  -,  when  x  =  1 ; 

£(a>)        x^-tf-x  +  l     0'  ' 

*'(*)  ~3^-2*-l~0'Wliena;"1' 

f"(x)         6x         3      ,  . 

•' ,,)  /  = =  -,  when  x  =  1. 

^"(a?)      6  a? -2      2' 

Hence  £  is  the  true  value  of  — — X —  when  a?  =  1. 

ar  — ar2  — a;  +  l 


SUCCESSIVE  DIFFERENTIATION  77 

PROBLEMS 

Find  the  true  values  of, 


x2  -Ux  +  24 


when  x  =  2. 


a2 -16         ,  , 

2.    when  #  =  4. 

a*  +  x  _  20 

0^  —  1 

3.  — when  x  =  l. 

x2  —  1 

4.  — _!__,  when  x  =  l. 

7x2-9x  +  2 

aJ4-2a8-t-2aj-l         ,  . 

6.  i+;;-^;iiiiBi.& 

z4-3ar}-7a;2  +  27a;-18 
7.   — ~r  when  a  =  1. 


„    x"  —  or     -, 

8.   when  x  —  a. 

x  —  a 

x-2 


(x  _  1)2  _  i 


when  x  =  2. 


10.    v     -r    v when  x  =  a. 

■y/x2—  a2 


11.    * —4-  when  a  =  a.     Here  Z-LEi  =  _  when  #  =  a  for 

all  values  of  n  and  the  true  value  of  the  ratio  cannot  be  found 
by  the  method  of  derivatives.  The  removal  of  the  factor 
(a  — »)»  from  numerator  and  denominator  leads  to  a  deter- 
minate result. 


CHAPTER   VI 

PARTIAL  DIFFERENTIATION  AND  INTEGRATION  OP 
ALGEBRAIC  FUNCTIONS 

Art.  27.  —  Partial  Differentiation 


If  to  pairs  of  arbitrarily  assigned  values  of  x  and  y  there 
correspond  one  or  more  determinate  values  of  z,  z  is  called  a 
function  of  the  two  independent  variables  x  and  y.  This  is 
denoted  by  writing  z  =f(x,  y). 

The  function  z  =  f(x,  y)  is  said  to  be  continuous  at  x0,  yQ  if 
limit  f(x0  -f  8X,  y0  +  82)  —  f(xo,  2/o)  =  0  when  limit  8j  =  0  and 
limit  82  =  0.  It  follows  that  if  f(x,  y)  is  a  continuous  func- 
tion of  x  and  y,  it  is  also  a  continuous  function  of  x  and  y 

separately.      The    converse    is 
not  necessarily  true. 

The  equation  z  =  f(x,  y), 
where  x  and  y  are  independent 
variables  and  z  is  a  continuous 
function  of  both  x  and  y,  when 
interpreted  in  rectangular  space 
coordinates  represents  a  curved 
surface.  Let  (x0,  y0,  zQ)  be  any 
point  in  the  surface.     If  x  re- 


— X 


Fig.  29. 


tains    the    fixed   value 


the 


equation  z  =f(x0,  y)  represents  the  projection  on  the  ZF-plane 
of  the   intersection    of    the    plane   x  =  x0  with    the   surface 

78 


PARTIAL  DIFFERENTIATION  79 

z  =  f(x,  y) .      If  z  =  f(x0,  y)  is  differentiated  with  respect  to 

y,  the  result  is  called  the  partial  derivative  of  z  =f(x,  y)  with 

dz 
respect  to  y,  and  is  denoted  by  the  symbol  — .     Denoting  by 

dz  dz  y  dz 

— 9  the  value  of  —  when  x  =  x0,  y  =  y0,  z  =  z0,  — ^  is  the  slope 

of  the  tangent  to  z  =f(x0,  y)  at  (y0,  z0)  and  measures  the  rate 
of  change  at  (x0,  y0,  z0)  of  z  when  the  point  (x,  y,  z)  moves 
along  the  curve  of  intersection  of  x  =  x0  and  z=f(x,  y). 

If  y  retains  the  fixed  value  y0,  the  equation  z  =f(x,  y0)  rep- 
resents the  projection  on  the  ZX-plane  of  the  intersection  of 
the  plane  y  =  yQ  with  the  surface  z  =f(x,  y).  If  z  =f(x,  yQ) 
is  differentiated  with  respect  to  x,  the  result  is  called  the 
partial  derivative  of  z  —  f(x,  y)  with  respect  to  x,  and  is  de- 

f/2  f}%  f)% 

noted  by  the  symbol  — .      Denoting  by  — -   the  value  of   — 
dx        ~  dx0  dx 

when  x  =  xQ,  y  =  y0,  z  =  zQ,  — -  is  the  slope  of  the  tangent  to 

dx0 

z=f(x}  y0)  at  (^0,  x0)  and  measures  the  rate  of  change  at 
(x0,  y0,  z0)  of  z  when  the  point  (x,  y,  z)  moves  along  the  curve 
of  intersection  of  y  =  yQ  and  z  =f(x,  y)* 

The  equations  of  the  straight  line  tangent  to  the  curve  of 

intersection  of  the  plane  x  =  x0  and  the  surface  z  =f(x,  y)  at 

dz 
(x0,  y0,  z0)  are  x  =  x0  and  z  —  z0  =  — "  (y  —  yQ)  ;  the  equations  of 

dyo 

the  straight  line  tangent  to  the  curve  of  intersection  of  y  —  y0 

dz 
and  z  =f(x,  y)  at  (x0,  y0,  z0)  are  y  =  yQ  and  z  —  z0  =  —^(x  —  x0). 

dyo 

The  plane  containing  these  two  tangent  lines  is  the  tangent 

plane  to  the  surface  z=f(x,  y)  at  (xQ,  y0,  z0).     In  the  analytic 

geometry  of   three   dimensions   it   is   proved   that  the   plane 

A  (x  —  x0)  +  B  (y  —  2/0)  +  G  (z  —  z0)  =  0    through    the   point 

(xo>  2/oj  zq)  contains  the  line  x  —  x0  =  a(z  —  z0)y  —  yQ=b(z  —  z0), 

*The  use  of  d  to  denote  partial  differentiation  was  introduced  by 
Jacobi  (1811-1851). 


80         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

when  Aa  -f-  Bb  +  C  =  0.      Hence  the  equation  of  the  tangent 

plane  is  found  to  be  z  —  z()=  —(x  —  x0)  H (y  —  ?/0). 

d#o  dy0 

Denoting  by  Z  the  angle  made  by  the  tangent  plane  with 
the  XY^plane,  cos  Z  = 


V        dx02     dy02J 


The  normal  to  the  curved  surface  z  =f(x,  y)  at  (x0,  y0,  z0)  is 
the  line  through  (x0,  y0,  z0)  perpendicular  to  the  tangent  plane 
at  this  point.     Hence  the  equations  of  the  normal  are 

X-X0=-^(z-Z0),  y-y0=-^(z-Z0). 
ox0  dy0 

For  example,  let  it  be  required  to  find  the  equation  of  the 
tangent  plane  and  the  equations  of  the  normal  to  the  sphere 

a2  +  V2  +  z2  =  14  at  (1,  2,  3).    Differentiating,  regarding  y  con- 

dz 
stant,     x-{-z—  =  0.      Differentiating,   regarding    x    constant, 

dz 
y  -\-z—  =  0.  At  the  point  of  tangency 

*.],  2/0  =  2,  ,0  =  3,  §&  =  -i  f&--i 

dz0         3    52/o         3 

Hence  the  equation  of  the  tangent  plane  is 

z-3  =  -Kz-l)-fG/-2), 

reducing  to  x  +  2  y  +  3  z  =  14 ;  the  equations  of  the  normal  are 
x  —  1  =  £  (z  —  3),  ?/  —  2  =  |  (z  —  3),  reducing  to  x  =  i  z,  2/  =  J  z. 
If  the  altitude  of  a  right  circular  cone  is  y  and  the  radius  of 
its  base  is  x,  denoting  the  volume  by  V,  V=^nx2 >y.  The 
variables  x  and  y  are  independent,  for  a  change  in  x  does  not 
cause  a  change  in  y.  Suppose  the  radius  to  remain  unchanged 
while  the  altitude  varies.    Differentiating  partially  with  respect 


PARTIAL  DIFFERENTIATION  81 

to  y.  — =\7rx2.     This  shows  that  if  the  base  of  the  cone 
dy 

remains  constant  the  volume  changes  ^ttx2  times  as  fast  as 
the  altitude.     If  the  altitude  remains  constant  and  the  base 

«J  yr 

changes,  —  =  §  wxy.    That  is,  the  volume  changes  §  irxy  times 
ox 

as  fast  as  the  radius  of  the  base. 


PROBLEMS 

In  the  following  functions  x  and  y  are  independent  variables. 

Form  —   and  — . 
dx  dy 

1.  z  =  xy. 

2.  z  =  x2  +  y2. 

3.  z=2xy2+5x?y+6x-8y. 

4.  z  =  (x2  +  s/2)!. 

5.  *  =  * 

2/ 

11.  Find  the  equation  of  the  plane  tangent  to 

x2-±y2  +  2z2  =  6  at  (2,-2,3). 

12.  Find  the  equation  of  the  normal  to  x2  —  4?/2  +  2z2=  6 
at  (2,  2,  3). 

13.  Denote  the  base  of  a  triangle  by  x,  its  altitude  by  y. 
Find  the  rate  of  change  of  area  when  the  base  remains 
unchanged. 

14.  The  pressure  of  a  gas  on  the  containing  vessel  varies 
directly  as  the  temperature  and  inversely  as  the  volume. 
That  is,  denoting  pressure  by  p,  temperature  by  t,  and  volume 
by  v,  p  =  c-,  where  c  is  a  constant.     Find  the  rate  of  change 


6. 

1  +x 
1-2/ 

7. 

x  —  y 

z  = ~. 

x-\-y 

8. 

x2  +  y2  +  z2=r2. 

9. 

z  =  ax2  +  by2. 

0. 

v1 +y2  4-*4-l 

a?     b2      c2 

82  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

of  pressure  when  the  volume  is  constant.  Also  the  rate  of 
change  of  pressure  when  the  temperature  is  constant. 

Art.  28. — Partial  Integration 

dz 
Example.  —  Find  z  by  integration  when  — -  =  3  yx  -j-  x?  -f-  y2. 

dx 

dz 
Since  y  is  considered  constant  in  forming  — ,  terms  of  z  con- 

dx  d> 

taining  only  y  and  constants  have  no  effect  on  the  value  of  — . 

dz  dx 

Hence  the  integration  of  — ,  considering  y  constant,  gives  all 
dx 

the  terms  of  z  which  contain  x,  but  it  is  necessary  to  add  an 
arbitrary  function  of  y,  f(y),  to  cover  the  terms  of  z  which  do 
not  contain  x.     This  integration  gives 

*  =  f  afy  +  i«4  +  y2%  +f(y)- 

If  the  value  of  z  is  known  for  some  special  value  of  x,  the 
value  of  f(y)  becomes  known  and  the  result  is  determinate. 
For  example,  if  z  =  3  y  +  5  when  x  =  0,  f(y)  =  3  y  +  5  and 
z  =  f  afy  +  \x*  +  y*x  +  3y  +  5. 


ntegrate, 
dx     J 

PROBLEMS 

3.    ^-  =  Xy2-Sx. 
dy 

dz          o 

.   —  =  xy£  - 

dx       U 

-3a;. 

a        OZ             2           .2 

4.    —  =  y2  —  X2. 
dy 

5. 

A. 

dx 

■-x  +  \y\ 

dz 

6.  —  =  Sx2  —  5y,  knowing  that  z  =  3 y2  -  7  when  x  =  0. 
dx 

dz 

7.  z—  =  2  y2  —  x,  knowing  that  z2  =  10  +  t/5  when  a  =  2. 


PARTIAL  DIFFERENTIATION  83 

Art.  29.  —  Differentiation  of  Implicit  Functions 

Represent  by  Ax  and  Ay  the  corresponding  changes  in  the 
values  of  x  and  y  when  y  is  a  continuous  implicit  function  of  x 
denned  by  the  equation  f(x,  y)  =  0. 

By  hypothesis  f(x,  y)  =  0  and  /(as  +  Ax,  y  +  Ay)  =  0.  By 
subtraction  f(x  +  Ax,  y  +  Ay)  —f(x,  y)  =  0,  whence,  by  adding 
and  subtracting  f(x,  y  -f-  Ay), 

[/(a?  +  Ax,  y  +  Ay)  -/(»,  y  +  Ay)] 

+  C/(^y  +  Ay)-/(aj,y)].=  0. 

Dividing  the  last  equation  by  Aa;  and  then  multiplying  numer- 
ator and  denominator  of  the  second  term  by  Ay,  there  results 

(1\  /(«?  +  Aa;,  y  +  Ay)  -fix,  y  +  Ay) 

V  }  Ax 

■  /O,  y  +  Ay)  -/(a?,  y)  Ay  _  Q 
Ay  Aa; 

Now  since  by  hypothesis  y  is  a  continuous  function  of  x, 

f(x  +  Ax,  y  +  Ay)  -f(x,  y  +  Ay) 
Aa; 

=  f(x  +  Ag,  y)  -/(a?,  y)  ± 
Aa; 

where  c  has  zero  for  its  limit  when  limit  Aa;  =  0.     Hence  when 
limit  Aa;  =  0,  (1)  becomes 

d/Q,  y) 

a/foy)  ■  df(x,y)dy  =  Q  and  *y=       dx 

dx  dy      dx  dx         df(x,  y) 

dy 
Eepresenting  f(x,  y)  by  u,  this  result  becomes 

du 
dy_  _^£# 
dx~      du 
dy 


84  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Denoting  by  —  and  —  the  values  of  the  partial  derivatives 
3  a  dx0  dy0 

—  and  —  when  x  —  x0,  y  =  y0,  the  equation  of  the  tangent  to 
dx  By 

the  curve  whose  equation  is  u  =/(x,  y)  =  0  at  the  point  (x0,  y0) 

fj/i  fill 

is  (y  —  ?/0)  — 9  -f  (x  —  x0)  — -  =  0  ;  the  equation  of  the  normal  is 
dy0  dx0 

e-*>t-<»-*>£=°- 

Let  z  be  a  continuous  implicit  function  of  the  independent 
variables  x  and  y  defined  by  the  equation  f(x,  y,  z)  =  0. 
Denoting  by  Ax  and  Az  the  corresponding  changes  in  the 
values  of  x  and  z  when  y  remains  unchanged, 

/(x  +  Ax,  y,  z  +  Az)  -/(»,  y,  z)  =  0, 
whence 

f(x  +  Ax,  y,  z  +  Az)  —/(a,  y,  g  +  Az) 

Ax 

■  /O,  ?/,  g  +  Az)  -/(a;,  y,  z)  Az  =  Q 

Az  Ax 

When  limit  Ax  =  0,  this  equation  becomes 

3/(x,  y,  Z)  +  3/(x,  y,  Z)  gz  =  0    whence  gz  = dx 

dx  Bz        dx  dx          df(x,  y,  z) 

dz 

In  like  manner,  denoting  by  Ay  and  Az  the  corresponding 
changes  in  y  and  z  when  x  remains  unchanged,  and  passing  to 
the  limit  when  limit  Ay  =  0, 

df(x,  y,  z) 
dz_  =  dy 

dy         df(x,  y,  z) 
dz 


PARTIAL  DIFFERENTIATION  85 

Representing  f(x,  y,  z)  by  w,  these  results  become 
du  du 

3£  =  _^     and      £— ft. 

dx  ou  dy  ou 

dz  dz 

Denoting   by  — ,  — -,   and  — -  the  values   of   the   partial 
dx0    dy0  dz0 

j     .      ,.         du    du   du      ,  ,,  ,. 

derivatives  — -,  — -,  — -  when  x  =  x0,  y  =  y0,  z  =  z0,  the  equation 
ox    oy  dz 

of    the    tangent    plane    to    the    surface   whose    equation  is 
u=f(x,  y,  z)  =  Q  at  the  point  (x0,  y0,  z0)  is 

the  equations  of  the  normal  are 

(*-*0)g-(*-*o)f;=o,  (y-^g-(,_^=o. 

Denoting  by  0X,  6y,  0Z  the  angles  made  by  the  normal  with 
the  rectangular  coordinate  axes  X,  Y,  Z  respectively, 

du0 
/i  dx0 

cos  ex = - p 

(du{?      dul      duly 
\dx02      dy02      dz02  J 

with  corresponding  values  for  cos  9y,  cos  0Z- 

Example.  —  Find  equations  of  tangent  plane  and  of  normal 
to  |  +  f  +  z2  =  l  at  (2,l,*Vll). 

Writing_f  +  ^,-l  =  0,£4^g  =  ||=2, 

Hence  when 

o  -i  ,     /re    5^o_4     Buq_1    dn0_1    /tt 

xQ  =  2,  2/0  =  l,  *b  =  *Vll,   5^-9'    ^"2,  ^~^ 


86  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  equation  of  the  tangent  plane  is 

£(*-2)+K2/-i)+ivn(z-ivn)  =  o. 

The  equations  of  the  normal  are 

and  jVll(2,-l)-K3-iVlI)  =  0. 

For  this  normal 

cos  0,  =  . 3233,  0X  =  71°8';  cos  0,  =  . 3638,  0y  =  68°2O'; 
cos  0Z  =  . 8341,  0,  =  33°29'. 

PROBLEMS 

1.  Find  equations  of  tangent  and  normal  to 

y3  —  3  xy  +  Xs  =  0  at  (x0,  y0). 

2.  Find  equation  of  tangent  plane  and  normal  to 

x2  +  y2-z2  =  0  at  (3,4,5). 

3.  Find  angles  made  by  normal  to  x2  4-  y2  —  z2  =  0  at  (2, 0, 4) 
with  the  coordinate  axes. 

4.  Find  equation  of  tangent  plane  to  xyz  =  8  at  (2,  2,  2). 

5.  Find  angle  made  by  plane  tangent  to  z  =  2  a,-2  4-  4  y2  at 
(2,  1,  12)  with  Xr-plane. 

Art.  30.  —  Successive  Partial  Differentiation  and 
Integration 

If  z=f(x,  y),  the  partial  derivatives  of  z  with  respect  to 

x  and  y,  —  and  — ,  are  in  general  functions  of  x  and  y.     The 

dx        dy  dz  . 

partial  derivative  with  respect  to  a;  of  —  is  called  the  second 

partial  derivative  of  z  with  respect  to  x,  and  is  denoted  by  the 


PARTIAL  DIFFERENTIATION  87 

symbol  — ,.      Hence  by  definition  — -  =  —  — -.      The   partial 
dx2  dx2      dx  dx 

dz 
derivative  with  respect  to  y  of  — -  is  called  the  second  partial 

ox 

derivative  of  z  with  respect  to  x  and  y,  and  is  denoted  by  the 

symbol  .     Hence  by  definition  == .     By  a  like 

by  ox  dydx      dy  dx 

.   ,.        d2z        d   dz        d2z         d   dz         A  ,  £    ,,. 

notation  — = ,    = .      An    extension    of    this 

dy2      dy  dy     dxdy      dx  dy 

.   .  •         .        d3z       d  d2z  d3z        d  d2z        d3z          d     d2z 

notation  gives  — -  = , = = , 

dx3      dx  dx2  dydxr      dydx-'  dxdydx      dxdydx 
and  so  on. 

For  example,  if  z  =  x3  -f  3  xy2  -f  yx2,   —  =  6  x2  -j-  3  y2  +  2  yx, 

dx 

P:  =  12x  +  2y,  J?L.=s6y+2x,  —=6xy+x2,  —=6y+2x. 
dx"  dydx  dy  dxdy 

d2z         d2z 

In  this  example  =  — — .     It  is  to  be  proved  that  this  is 

dxdy      dydx 

always  true.     The  proposition  may  be  formulated  thus: 

If    the    function  z  =  f(x,  y)    and    its    partial    derivatives 

dz      dz       d  z         d2z 

— ,    — -,    ,    have   determinate   finite   values,   and  z, 

dx     dy    dxdy     dydx 

dz      dz  d  z  b2z 

— ,    —   are  continuous  functions  of  x  and  y, 


dx     dy  dydx      dxdy 

By  hypothesis   S1  =  f{x  + l^x,  y)  -  f(x,y)  ±      where  f  de_ 
dx  Ax 

pends  on  x,  y,  and  Ax,  and  approaches  zero  with  Ax.     It  is 

convenient  to  write  e  =  e1f1(x,  y),  where  ^  depends  only  on 
Ax  and  approaches  zero  with  Ax  while  fx  (x,  y)  must  be  finite. 
Similarly 

d2z  _f(x-\-Ax,  y  +  Ay)  —  f(x,  y  -f  Ay)—f(x  -f  Ax,  y)  +f(x,  y) 
dy  dx  Ax  Ay 

±  €M*,y  +  ±y)-M*,y)  ±  ^  {Xy  y)i 

Ay 

where  limit  /i(^  .V  +  A.v) -/ife  ^  wiien  limit  Ay  =  0,  must 
Ay 


88  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

be  finite,  e2  must  vanish  with  Ay,  and  fa  (x,  y)  must  be  finite  by 
hypothesis. 

In  like  manner  —  =  f^  y  +  Ay)  ^I^ill  ±  Hffa  y\  ana 
by  Ay 

d2z  =     0  +  Ax,  y  +  Ay)  -  f(x  +  Ax,  y)-f(x,  y+Ay)+f(x,  y) 
dxdy  Ay  Ax 

where    c3    must  vanish   with   Ay,   e4  must  vanish  with   Aa;, 

limit  f*(x  +  *x>y)-f*(x>y)  when  limit  Aa?  =  0  must  be  finite, 
Ax 

and  fA(x,  y)  must  be  finite.     It  follows  that  when  limit  Ax  =  0 
and  limit  Ay  —  0, 

limit  /(>  +  As,  y  +  Ay)  -/(a;  +  Aa;,  y)  -/(a;,  y  +  Ay)  +  /(a,  y) 

AyA# 
dh         dh 


dxdy      dydx 

PROBLEMS 

Form  —    —       d*z      —    —       d*z     of 
dx     dx*    dydx     dy     dy2'    dxdy 

1.   z  =  a2+y2.      2.   «  =  a^y.       3.   z  =  ^±^.      4.  2  =  #(y-2). 


5.   2  = 


a+y 


6.  2  =  #2y2.      7.2  =  a?'y*.      8.  2  =  x*y~K 


d2z 
Integrate    9.   —  =  x?y  +  3  x  —  5  y  +  2.      Integrating,  con- 

dz 
sidering  y  constant,  — -  =  |  ar*y  +  §  a?2  —  5  ya?  +  2  a;  +  /x  (y). 
oa? 

Integrating  again,  considering  y  constant, 

2  =  TV  x*y  f  J  x*  -  I  yx>  +  x2  +  /1  (y)  •  *  +  /,  (y). 

The  arbitrary  functions  /,(y)  and  /2(y)  become  known  if  2  and 

—  are  known  for  some  value  of  x.     Suppose  that  when  x  =  0, 
dx 


PA  R  TIAL   BIFFEREN TIA  T10N 


89 


«  =  V  +  3,  ||=  2/2  +  5.       Then  /&)  =  ?/2  +  5,    /8(y)  =  y  +  3, 
and  the  result  becomes 

z  =  tV  x*y  +  i^  + 1 2/^  +  v?  +  ^2/2  +  5  *  +2/  +  3. 
d2z 


10. 


dxdy 


xy  +  2x  —  7  y  -\-  5,   knowing  that,   when  x  =  0, 


dz 


z  =  y  —  2  and  when  y  =  0,  —  =2x?. 

dx 


11. 


a*? 


14. 


dydx 


3xy2. 


12. 


d2z 


5a;6?/ 


=  (*_3)(y-2). 


as^-2.        13.  ||=^_/. 

15-  ;nr=a,&-5)- 

5?/  dx 


Art.  31.  —  Area  of  Any  Curved  Surface 

Let  the  equation  of  the  curved  surface  be  z=f(x,  y)  con- 
tinuous in  x  and  y.  Planes  perpendicular  to  the  X-axis  at 
intervals  of  Ax  divide  the  given  surface  into  strips  of  surface 
CDEF.  Planes  perpendicular  to  the  F-axis  at  intervals  of 
Ay  divide   these  strips  into  elements  of  surface  abed.     The 

y  =  BG 

strip  CDEF  =  2   abed,  and  the  given  surface 

y  =  0 
x=OA  x=OA  y  —  BG 

4  =  2  CDEF  =  2      2  abed. 

as=0  x  =  0      y  =  0 

This  summation  holds  what- 
ever be  the  magnitude  of  Ax 
and  Ay.  The  projection  of 
the  element  of  surface  abed 
on  the  XF-plane  is  the  rec- 
tangle ajbtfidn  whose  area  is 
Ax  Ay.  The  plane  through  ad 
and  the  line  ab'  parallel  to 
afa  intersects  the  prism  which 


Fig.  80. 


90 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


projects  abed  on  the  .XF-plane  in  the  parallelogram  ab'c'd. 
Denoting  by  6'2  the  angle  made  by  the  plane  of  ab'c'd  with 

the  XF-plane,  area  ab'c'd  =        J  «      When   Ax    and    Ay  ap- 

cos  6  e 

proach  zero,  the  plane  of  the  parallelogram  ab'c'd  approaches 
the  tangent  plane  to  the  curved  surface  at  a,  and  the  paral- 
lelogram ab'c'd  approaches  the  surface  element  abed.  Hence, 
when  limit  Ax  =  0  and  limit  Ay  =  0, 


x=OA  y  =  BC  x=OA  y  =  BG 

A  =  2      2    abed  =  limit  2       2   ab'c'd 

x  =  0      y  =  0  x  —  0      y=0 


x=OA  y  =  BG 


limit  2 


A.tAi/ 


y  =  0     COS0'a 

Example.  —  Find  the  area  of  the  groin  formed  by  the  inter- 
section of  the  semicircular  cylinders  x2  -J-  z2  =  r2,  y2  +z2  =  r2. 

One-eighth  of  the  surface  of  the  groin  is  bounded  by  the 

cylinder  x2+z2=r*,  the  ZX-plane, 
the  Xy-plane,  and  the  intersec- 
tion of  the  cylinders,  which 
intersection  lies  in  the  plane 
y—x.  Hence  for  this  part  of 
the  surface  of  the  groin 

dz_ _  _x     dz 
dx  z '    dy 

and 


0, 


Fig.  31. 


8/ 


Jf*x  =  r    S*y 
i  =  0    Jy  = 


=  0    (r 


dxdy     __       /**=r 

l  —  x2)^  J*=0    I 


xdx 


(r2-  xrf 


=  _8rj(r2-ayJ)*+C 


8r 


PARTIAL  DIFFERENTIATION 


91 


Art.  32.  —  Volume  of  Any  Solid 

Denote  by  V  the  volume  of  the  solid  bounded  by  the  con- 
tinuous surface  z=f(x,y)  and  the  coordinate  planes  IT, 
XZ,  YZ.  Planes  perpendicular 
to  the  X-axis  at  intervals  Aa* 
divide  the  solid  into  laminae 
LL';  planes  perpendicular  to 
the  Y"-axis  at  intervals  Ay  divide 
these  laminae  into  prisms  PP*\ 
planes  perpendicular  to  the  Z-axis 
at  intervals  Az  divide  these 
prisms  into  elements  of  volume 
aa'.     The  volume  of 


aa' 


Aa?  •  Ay  •  Az ; 


the  volume  of  PP'  =  2    Aa;  •  Ay  •  Az  —  0  •  A#  •  Ay  •  Az,  where  0 
is  less  than  unity.     When  Aa*,  Ay,  Az  approach  zero,  volume 

x  =  PD 

of  PP'  =  2   dX'dy'dz  —  6-dX'dy-  dz.    Since  6  •  dx  •  dy  >dz  is  an 

z  =  PD 

infinitesimal  of  a  higher  order  than  2    dx  -dy  >dz  =  PD  •  dx  •  dy, 

z  =  0 

dX'dy  •  dz. 

y  =  BC 

The  volume  of  the  lamina  LL'  =  2   PP' ;  the  volume  of  the 


solid  is 


y  =  0 

V=2  LL'=5      2   PP'. 

x=0  x=0       y=0 


Hence  in  the  limit 


J  (  dxdydz. 

x=0        c/tf  =  0        i/?  =  0 


92 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Example.  —  Find  the  volume  of  the  solid  bounded  by  the 

ZX-plane,  the  XF-plane,  and 
the  planes  x=a,y=b,  z—mx. 
In  this  problem 

J<+x  =  a    S*y  =  b    S»z  =  mx  , 

clxdydz 


S»x  =  a    /-*y  =  b 

=  m  I  I       xdxdy 

c/x  =  0    */y  =  0 
/»x  =  a 

—  mb  |        xdx  = 

Jx  =  0 


\mbar. 


Art.  33.  —  Total  Differentials 

Let  u  =f(x,  y)  represent  a  continuous  function  of  the  inde- 
pendent variables  x  and  y.  Let  Axu  denote  the  change  in  the 
value  of  u  corresponding  to  a  change  of  Ax  in  the  value  of  x 
when  y  remains  unchanged;  Ayu  the  change  in  the  value  of  u 
corresponding  to  a  change  of  Ay  in  the  value  of  y  when 
x  remains  unchanged :  Au  the  change  in  the  value  of  u  corre- 
sponding to  a  change  of  both  x  and  y  by  Ax  and  Ay  respectively. 
Then 
(1)  A^  =  ^  +  A^>-/^>Att, 


(2) 


Ax 

Ay  * 

(3)     Au  =f(x  +  Ax,y  +  Ay)  -f(x,  y) 

= f(x  +  ^x>  y  +  A-y)  -f(x>  y  +  Ay)  Aaj 

Ax 

f(x,y  +  Ay)-f(x,y) 

Ay  y 

=  /(a?  +  AX,  y)  -/(a,  y)  ±  e  ^ 
Ax 


+  /fo  y  +  Ay)-/(gty)  A 
Ay 


PARTIAL   DIFFERENTIATION  93 

where  c  vanishes  when  Ay  approaches  zero.  If  Ax  and  Ay  are 
indefinitely  decreased,  becoming  dx  and  dy,  and  the  correspond- 
ing values  of  Axu,  Ayu,  Au  are  denoted  by  dxu,  dyu,  du  respec- 
tively, equations  (1),  (2),  and  (3)  become 

dxu  =  ~dx,  dyu  =  -^dy,  du  =  -—  dx  «f — dy  =  dmu  +  d9u. 

ox  ay  ox  ay 

The  quantity  dxu  is  called  the  partial  differential  of  u  with 
respect  to  x ;  dyu  the  partial  differential  of  u  with  respect  to  y ; 
du  the  total  differential  of  u.  The  equation  du  =  dxit  +  cZyw 
expresses  the  fact  that  the  total  differential  equals  the  sum  of 
the  partial  differentials. 
For  example,  if 

u  =  x2  +  y2  —  xy,  dxu  =  —  dx  =  (2  x  —  y)  dx ; 
ax 

dyu  =  -^dy  =  (2y-  x) dy, 
dy 

du  =  ^dx  +  ^dy  =  (2x-y)dx  +  (2y-x)  dy. 
ox  ay 

The  differential  expression  (4)  du  =  Pdx  -f-  Qdy,  where  P 
and  Q  are  functions  of  the  independent  variables  x  and  y,  is 
said  to  be  exact  if  it  can  be  obtained  by  differentiating  some 
function    (5)    u=f(x,y).       The    differential    of    (5)    is    (6) 

du  =  ^dx  +  —  <fy,and  if  (6)  is  identical  with  (4),  (J)—  =  P, 
ox  ay  ax 

(8)  f*  =  Q.     The  partial  ^derivative  of  (7)  is  -^-  =  —  :    the 
dy  aydx      dy 

partial  o-derivative  of  (8)  is  ^-=  |S.     Since  J!lL  =  J!lL 

oxoy      ox  aydx      dxdy 

the  hypothesis  that  (4)  is  exact  leads  to  the  condition  —  =  —-*. 

dp  so  .  d,J    Sx 

Conversely,     if       —  =  — -,       the     differential     expression 
dy       ox 

du  =  Pdx  +  Qdy  can  be  integrated.     All  the  terms  of  u  which 


d_ 

dx 


94  DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

contain  x  are  found  by  integrating  Pdx,  considering  y  constant. 
That  is,  (1)  ?*  =  j  Pdx  +  Y,  where  Y  stands  for  the  terms  of 
u  which  do  not  contain  x.  To  find  Y  from  the  ^/-derivative  of 
(1),  —-  =  —  I  Pdx  +  — — .     If  the  expression  du  =  Pdx  -f  Qdy 

can  be  integrated,  -^=Q.      Hence  —  |  Pcto  H =  Q,  and 

**_*    a  r     dy  dyJ        dy 

~t~  —  H  ~  ~q~  I  Pdx.     Y  can    be    found    by    integration     if 

Q  —  — -  I  Pcfcc  is  independent  of  #.     Forming  the  ^derivative 
dyJ 

of  this  expression, 

\         ay  J         J       ox      ox  oyj 

da;       di/  da\/  d#       6?/ 

by  hypothesis.     Hence  Y"  can  be  found  by  integration  and  the 
integral  of  the  given  differential  expression  becomes  known. 
Consider,  for  example,  the  differential  expression 

du  =  (Sxy2  -  x2)  dx  -  (1  +  6y2  -  3a%)  dy. 

This  can  be  integrated  since  —  =  6  aw  =  — ^.     Integrating  the 

dy  dx 

first  term  of  dw,  considering  y  constant,  u  =  $  a%2  —  A  a?  +  F. 
Differentiating  partially  with  respect  to  ?/, 

Whence         1I=_1_62/2  and  r=-y-2y8+(7. 
Finally  w  =  fafy2  -  -^ar5  -  ?/  -  2  y3  +  O. 


PARTIAL  DIFFERENTIATION  95 


PROBLEMS 

Form  the  total  differential  of 

x  o  x2 

1.    u  =  xy.  2.    u=-.  3.    u  =  x*y.  4.    w  =  — . 

y  y 

5.  The  pressure  of  a  gas  on  the  containing  vessel  varies 
directly  as   the   temperature   and   inversely  as   the   volume; 

that  is,  p  =  c-,  where  c  is  a  constant.     Find  the  change  of 

v 

pressure  when  the  temperature  remains  constant;  the  change 
of  pressure  when  the  volume  remains  constant ;  the  change  of 
pressure  when  temperature  and  volume  both  change. 
Integrate  the  differential  expressions : 

6.  du  =  (3x2  +  2ax)dx+  (ax2 +  3y*)dy. 

7.  du=(x*  +  3xy2)dx  +  (y3  +  3x2y)dy. 

8.  du  =  (x2  —  4 xy  —  2 y2)dx  +  (y2  —  kxy  —  'lx2) dy. 

9.  u  =  ax*yz.     Show  that  x—  -f-  ?/—-'=  5  w. 

ax        dy 

10.  u  =  ax2?/*  +  bxy\     Show  that  x—  +  w—=5w. 

ox         dy 

11.  u  =  F(x,  y),  where  F(x,  y)  is  homogeneous  of  degree  n. 

Show  that  x— - -\-y-—  =  ri'U. 
ox         oy 

The  terms  of  F(x,  y)  are  of  the  general  form  ur  =  Ax(n~r)yr. 

For  all  terms  of  this  form  #— - -  +  y~  =  n  •  wr,  and  the  truth 

ox         oy 

of  the  proposition  follows  directly.     This  is  Euler's  theorem 
on  homogeneous  functions. 

12 .  If  du  =  fx  (x,  y)  dx  -f-  f2  (x,  y)  dy  is  exact  and  homogeneous 
of  order  » •—  1,  show  that  n  •  u  =  xfx  (x,  y)  -\-  yf2  (x,  y)  +  G. 

Denoting  the  integral  of  the  given  differential  expression 
by  u  =  f(x,  y),  where  f(x,  y)  must  be  homogeneous  of  degree 


96         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

n,  since  differentiation  diminishes  by  unity  the  degree  of  an 

expression,  du  =  -^dx  -f  -^dy  and  n  •  u  =  x—  +  y — •     In  the 
ox  dy   "  dx        dy 

fill  fj?/ 

given  expression  —  =  fx  (x,  y),   —  =  f2  (x,  y).     Hence 

n  •  u  =  xf^x,  y)  +  yf2{x,  y). 
Integrate  the  following  expressions : 

13.  du  =  (2y2x  +  3 y3) dx  +  (2 afy  +  9 a^2  +  8?^) dy. 

14.  du  =  (y2  +  6xy)dx+(2xy  +  3x2)dy. 

15 .  dw  =  (1  af"*y  +  6  y*)  c?»  +  (a?*  +  3  ic?/-^)  dy. 


Art.  34.  —  Differentiation  of  Indirect  Functions 

If  z  =f(x,  y)  and  y  —  F{x),  z  is  said  to  be  a  function  of  x 
directly  and  also  indirectly  through  y.  Denoting  by  Ax,  Ay, 
and  Az  the  corresponding  changes  in  the  values  of  x,  y,  and  z, 
Az=f(x  +  Ax,y  +  Ay)-f(x,y)  and  Ay  =  F(x  +  Ax)-F(x), 
whence 

Az_/(a;  +  Aa;,  y+Ay)-/(a;,y+Ay)      f(x,y+Ay)-f(x,y)Ay 
Ax  Ax  Ay  Ax 

and  —^  as     ^a?~i~ — ^^ — a2L      Passing    to    the   limit   when 

Ax  Ax 

limit  As  =  0,  ^  =  ^  +  |i^  and  &=*»(*). 

aa;     da;     oy  da;  da; 

Example.  —  A  point  moves  along  the  intersection  of  the 
paraboloid  z  =  3 x2  +  5 y2  and  the  plane  y  =  2x.     Find  the  rate 

of  change  of  2  and  x.     Here  —  =  6  a;,  —  =  10  y,  and  ^/  =  2. 

,  5a;  dy  dx 

Hence  —  =  6a;  +  20y. 

da; 


PARTIAL  DIFFERENTIATION 


97 


PROBLEMS 
dz 
Determine  —  when 
dx 

1.  x2  +  y2  +  z2  =  25  and  y  =  2x  +  3. 

2.  x2  +  y2  +  z2  =  0  and  2x-3y  =  0. 

3.  -2  +  £  +  -2  =  l  and^  +  ^r2. 
a2     62     c2 

4.  ^  +  |]  +  -'  =  1  and  tf-3xy  +  x*  =  0. 
a2     b2     c2 

5.  Determine  ■— -  and  ~  when  u  =  z3  and  z  =  x-\-y. 

ox  oy 


Art.  35.  —  Envelopes. 

The  equation  f(x,  y,a)  =  0  represents  a  curve  whatever  the 
value  of  a.  By  assigning  to  a  all  real  values  an  infinite  system 
of  curves  is  obtained.  The  locus  to  which  every  curve  of  this 
system  is  tangent  is  called  the  envelope  of  the  system  of 
curves.     The  equation  of  the  envelope  is  to  be  determined. 

Let  (1)  f(x,  y,a)  =  0  and  (2)  f(x,  y,  a  -f-  Aa)  =  0  represent 
any  two  curves  of  the  system.  Their  points  of  intersection 
approach  the  points  of 
tangency  of  f(x,  y,a)  =  0 
with  the  envelope  when  Act 
approaches  zero.  Hence 
the  envelope  may  also  be 
defined  as  the  locus  of  the 
ultimate  intersections   of 

f(x,  y,a)  =  0 

and     f(x,  y,a  +  Aa)  =  0 

when  Aa  approaches  zero. 


Fig.  84. 


The  points  of  intersection  of  (1)  and 


98         DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

(2)  satisfy  f(x,y,a)  =  0   and  £<&  &  "  +  Act)  -f^V,  «)  =  p, 

Aa 
These  equations,  when  Aa  approaches  zero,  become 

/(»,  y,a)  =  0  and  t-/(«,  2/>  «)  =  0. 

The  elimination  of  a  from  the  last  pair  of  equations  gives  the 
equation  of  the  envelope. 

In  like  manner  the  envelope  of  the  singly  infinite  system 
of  surfaces  f(x,  y,  z,a)  =  0   is  found  by  eliminating  a  from 

f(x>  V)  h  a)  =  0  and  ^-/fo  V,  *>  »)  =  0. 

The   envelope   of  the   doubly  infinite   system   of  surfaces 
f(x,  y,  z,  a,  b)  =  0    is    found    by   eliminating  a  and   b   from 

f(x,  y,  z,  a,  b)  =  0,     —  f{x,  y,  z,  a,  b)  =  0,    —  f(x,  y,  z,  a,  b)  =  0. 

Example.  —  Find  the  envelope  of  the  system  of  circles  with 
centers  on  the  X-axis  and  radii  one-third  of  the  distance  of 

center  from  the  origin. 

a2 
The  equation  of  the  system  of  circles  is  (x  —  a)2  +  y2  =  —, 

j 
where  a  is  the  distance  of  center  from  origin.     Differentiating 

with  respect  to  a,  —2(x  —  a)  =— .  Eliminating  a  from  the 
equations  (x  -  a)2  +  y2  =  |L  and  -2(x-a)  =  — ,  y2  =  \x*> 
the  equation  of  the  envelope.     This  equation  represents  the 

two  straight  lines  y  —  — —  x  and  y  = x. 

2V2  2V2 


PROBLEMS 

1.   Find  the  envelope  of    the  system  of    lines  ^  +  ^  =  1 
when  a  +  b  =  c,  where  c  is  a  fixed  constant. 


PARTIAL  DIFFERENTIATION  99 

2.  Show  that  the  envelope  of  the  normals  to  the  parabola 
if  =  2px  is  a  semi-cubic  parabola. 

3.  Find  the  envelope  of  the  system  of  ellipses  — (-^-  =  1 
for  which  the  area  nab  is  constant. 

2  2 

4.  Find  the  envelope  of  the  system  of  ellipses  — -f«7^-  =  l 


for  which  a+b  =  c,  where  c  is  a  fixed  constant. 


a2     52 


5.  Find  the  envelope  of  the  system  of  planes  —  -f-  —  -t— —  =  1 

when  abc  =  m,  where  m  is  a  fixed  constant. 

6.  Find  the  envelope  of  the  system  of  spheres  whose 
centers  lie  in  the  XF-plane  and  whose  radii  vary  as  the  dis- 
tance from  origin  to  center. 


CHAPTER   VII 


CIRCULAR  AND  INVERSE  CIRCULAR  FUNCTIONS 


Art.  36.  —  Differentiation  of  Circular  Functions 


Denote  by  x  the  circular  measure  of  an  angle  less  than  a 
right  angle.     From  the  figure  triangle   OPD  <  sector  OPA 

<  triangle  OTA.     Hence 

-J?*2 •  sin  a;  •  cos x  <  i r2 •  x  <  £ r2 •  tana; 
x      .     1 


and  cos#< 


< 


This  in- 


sin  x      cos  x 
equality  is  true  for  all  values  of  x 

less  than  -•     When  x  approaches 

zero,  cos  x  approaches  unity.    Since 

lies    between    two  numbers, 


x 


sin  a; 


cos  a;  and  ,  whose   common   limit    is   unity   when   x   ap- 


cosa; 

proaches  zero,  limit  -^—  =  1  when  limit  x  =  0.     This  limit  is 
sin  a; 

fundamental  in  this  chapter. 

Denote  by  u  a  continuous  function  of  x  which  changes  by 
Au  when  x  changes  by  Ax.     Then 

T     d               v    ..  sm(u  4-  Au)  —  sinw  Au 
I.   —  sin  it  =  limit  — *— -2- L 

dx  Au  Ax 

—  V -mit  *  S^n  J  —  '  C°S  ^  ^  -  ~  ^?^  •  — 

A  Aa; 

100 


CIRCULAR  FUNCTIONS  101 

.    Au 
sin 


,.  2  /         Au\  Au 

=  limit  — r cos   u  +  — —  ] 

Au  \  2  J  Ax 

2 

=  cos  u when  limit  Ax  =  0. 

dx 

dx  dx       \2       J  V       J  dx\2       J 

du 

sb—  siiim  •  — 
dx 

d    .  d 

cos  u  —  sm  u  —  sm  u  —  cos  u 
TTT     d  .  d  smu  dx  dx 

III.  —  tan  u    = = 

dx  dxcosu  cos'u 

2      du  .     •   o      du        du 

cosJ  i£ h  sir  w  •  —        — 

dx  dx        dx  9      du 

— =  — —  =  sec^u* — • 

cos2w  cos2w  dx 

IV.  Acotw    =Atan^-^  =  sec2^-wV—  (*-*} 
dx  dx        \2        J  \2       J  dx\2        J 

2      du 

=  —  cosec^M 

dx 

du 

■xt     d  d      1  ax 

V.  —secw    =  - = — 

ax  dxcosu        cos'u 

smu     1     du      .  M„      du 

=  tan  u  •  sec  u 


cos  u  cos  u  dx  dx 

dx  dx 


VI.  -^cosecu  =  -^-sec/7 


■»(l-)-"(l-)s(f-) 


=  —  cot  u  '  cosec  u 

dx 


102       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

VII.   —  vers  u     =  —  (1  —  cos  u)  =  sin  u  •  -^« 
(fcc  d#  dx 

VIII.  —  covers  w  =  —  (1  —  sin  w)  =  —  cos  u  •  —  • 
da;  da?  da; 

Example  I.  —  Differentiate  cos  (3  x2). 

—  008(3^)  =  -sin  (3  x2)  .  —  (3  a;2)  =  -  6  »  •  sin  (3  x2\ 
dx  dx  ■ 

Example  II.  —  Assuming  that  the  relative  rate  of  change 
of  cos  x  and  x  remains  the  same  as  at  30°  throughout  the  next 
10',  calculate  cos  30°  10', 

sin  30°  =  .5,  cos  30°  =  .86603,  and  —  cos  x  =  -  sin  x. 

dx 

Hence  at  30°  the  relative  rate  of  change  of  cos  x  and  x  is 
—  .5.  Since  x  is  the  circular  measure  of  the  angle,  it  is 
necessary  to  express  the  increment  of  the  angle  in  circular 
measure.       The    circular    measure    of    an    angle    of    10'    is 

f 14159  =  .0029088.     To  a  change  of  .0029088  in  the  value  of 

180  x  6 

x  there   corresponds   a   change    of  —  .0014544   in  the   value 

of    cos  x  under    the    assumption    of    the    problem.      Hence 

cos  30°  10'  =  .86458. 

^mtm  PROBLEMS 

Differentiate, 

1.  sin  (a?2).  8.  sinaj  •  cos(3#).  15.  cos* (5a?). 

2.  sin  (5  a).  9.  7  x  .  tan  (3  x) .  16.  tan  (a;*). 

3.  sin (7 a).  10.  cot2  (2 #).  17.  sin7 a. 

4.  2(sin#)2.  11.  sec  (x2).  18.  cos4 a,-. 

5.  sin  (3  a)2.  12.  covers  (3  x  -f  5).  19.  sin (3a?)*, 

6.  4sin/-\  13.  sin(2#  —  7).  20.  tan3a. 

7.  5cos2(2a).  14.  tan (3  — 4a;).  21.  cot5#. 


CIRCULAR  FUNCTIONS  103 

22.  Find  the  rate  of  change  of  sin  a;  when  x  =  45°. 

23.  Find  the  rate  of  change  of  tana;  when  x  =  45°. 

24.  Find  the  rate  of  change  of  cos  x  when  x  =  240°. 

25.  Find  when  sin  a;  changes  half  as  fast  as  x. 

26.  Find  when  cos  x  decreases  half  as  fast  as  x  increases. 

27.  Find  when  tana;  changes  four  times  as  fast  as  x. 

28.  Assuming  the  rate  of  change  of  sin  a;  to  remain  the  same 
as  at  60°  throughout  the  next  5',  calculate  sin  60°  5'. 

29.  Assuming  the  rate  of  change  of  tan  x  to  remain  the  same 
as  at  45°  throughout  the  next  2',  calculate  tan  45°  2'. 

30.  ®  =  (a2  -  x2)?  and  x  =  a .  sin  0.     Find  % 
dx  dO 

Since  dy  =  cly.<te  and  ^=  (a?  -  x2)?  =  acos<9,  ^=acos0, 
dO     dx  dd  dx     K  J  d6 

^  =  a2-cos20. 

dO 

31 .  %L  =  x  (x2  +  a2)*     Find  %  when  x  =  a-  tan  0. 
dx        K  }  dO 

32.  ^-=x2(x2-a2)^.     Find  ^  when  a  =  a.sec0. 
dx         K  J  dd 

33.  ^  =  a;2(2aa'-aM     Find  %  when  a;  =  a  (1  -  cos  0). 
dx         K  )  dd  v  } 

34.  Form  -^  and  — \  for  circle  x  =  B'COS0,  y=B-smO. 

dx  dx2 

Since  x  and  2/  are  continuous  functions  of  0,  if  Ax,  Ay,  and  A0 
denote  corresponding  changes  in  x,  y,  and  0,  limit  Aa?  =  0  and 
limit  Ay  =  0  when  limit  A6  =  0.     For  all  values  of  AO, 
Ay  dy 

*!Up.    Hence  when  limit  A0  =  0,   &  =  ^. 
Aa;     Aa;  dx     dx 

A0  d$ 


104       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

35.  Form     ^     and     ^     for    cycloid    x  =  R0-R$m0, 

dx  dx2 

y  =  E—  RcosO. 

36.  Form     ^     and     ^     for    cycloid     x  =  R-Rcos6, 

dx  dx2 

y  =  R6  +  RsmO. 

37.  Show  that  2  tan  x  —  tan2#  has  a  maximum  value  when 

4 

38.  Show  that  tan  x  +  3  cot  x  has  a  minimum  value  when 

39.  Find  the  maximum  radii  vectores  of  r  =  a  sin  (3  0). 

40.  Examine  sin  x  cos3 a;  for  maxima  and  minima. 

41.  Examine  since  +  cos  a;  for  maxima  and  minima. 


Fig.  36. 

42.  Find  the  length  of  the  arc  of  the  sector  which  must  be 
cut  from  a  circular  piece  of  sheet  iron  so  that  the  remainder 
may  form  a  conical  vessel  of  maximum  capacity. 

V  =  -r3  sin2#  cos  x. 
o 

43.  A  steamer  whose  speed  is  8  knots  per  hour  and  course 
due  north  sights  another  steamer  directly  ahead  whose  speed 
is  10  knots  and  course  due  west.     What  course  must  the  first 


CIRCULAR   FUNCTIONS  105 

steamer  take  to  cross  the  track  of  the  second  steamer  at  the 
least  possible  distance  from  her  ? 

Find  the  true  values  of, 

A.     1  —  cos  x     ,  A        .  „     x  —  sin  x  cos  x     -l  ^ 

44. when  x=0.       47. 


45.    x~smx  when  a  =  0.       48. 
ar 


46.   ^M  when  a>  =  0.  49. 


ar3 

tan  a?  —  a? 
a;  —  sin  x 

when  aj  = 

=  0. 

#sin#         , 

; —  when  x 

x  —  2  sin  a; 

=  0. 

50.    Show  that    the    radius   of    curvature  of    the  cycloid 
x  =  RO  —  R  sin  0,  y  =  R  —  R  cos  0  is  twice  the  normal. 


Art.  37.  —  Evaluation  of  the  Forms  oo  •  0,  co  —  oo,  — 

GO 

If  f(x)  •  <f>  (x)  when  x  =  a  takes  the  form  go  •  0, 
f{x)  .  +(»)  =!M  =  <>  when  s  =  a, 

and  the  true  value  of  the  expression  may  be  found  by  the 
method  of  Art.  26. 

If  f(x)  —  <f>  (x)  takes  the  form  co  —  oo  when  x  =  a, 

1  1 


f(x)-Hx)=^-^^^Kl^enx  =  a, 
f(x)      +@)       /(xj^j 


and  again  the  method  of  Art.  26  may  be  applied.     The  reduc- 
tion of  the   form  co  — 
effected  more  directly. 


tion  of  the   form  co  —  co  to  the  form  -  can  frequently  be 


106       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

1 

If  y==/M  =  ^  when  a  =  a,  y  =  ^  =  £  when  x  =  a,  and 

the  method  of  Art.  26  gives 

Hence,  y  =  y  ,;  j  and  w  =  Z^2  =  /  A  /  when  a;  =  a.  That 
is,  the  form  —  is  evaluated  in  precisely  the  same  manner  as 
the  form  -• 

Example  I.  —  Evaluate  (a2  —  x2)  tan—  when  x  =  a. 
v  ;        2a 

(a2  —  x2)  tan^—  =  0  •  oo  when  x  =  a. 

Z  a 

/  2        tv  i     7r#      a2  —  x2     0      t 

(a2  —  ar)  tan  —  = =  -  when  x  —  a. 

*       2<*      cot^     ° 
2a 

Hence,  (a2  —  x2)  ta,n~  = — =  —  when  x  =  a. 

2«      __ZLCOSec2^        * 
2a  2a 

Example  II.  —  Evaluate  sec  x  —  tan  x  when  x  =  -• 

2 

sec  x  —  tan  x  =  oo  —  oo  when  <c  =  -• 

2 

sec  a;  —  tan  x  =  — — =  -  when  #  =  -• 

cos  a;        0  2 

Hence,  sec  x  —  tan  a;  =  ~G0SX  —  o  when  a;  =  -• 
—  sin  x  2 


CIRCULAR  FUNCTIONS  107 


PROBLEMS 


Evaluate,   1 .   (1  —  x) tan  — -  when  seal, 


1 


7T 


2.   when  a;  =  0.  3.   x sin2-  when  a;  =  oo« 

cot^  x 

2 

4.   x  cot  a;  when  a;  =  0.  5.  sec  (3  a;)  cos  (5«)  when  x  =  -• 

6.  2  a;  tan  a;  —  7r  sec  a;  when  a;  = -• 

2 

7 .  (1  —  tan  x)  sec  (2  a;)  when  x  =  j- 

Art.  38.  —  Integration  of  Circular  Functions 
The  eight  formulas  of  Art.  36  may  be  written, 

I.    I  cos  u  •  —  =  sin  u  +  (7,        III.    |  sec2  w  •  —  =  tan  u  -f  (7, 
*/  da;  J  dx 

II.    I  sinw-  —  =  —  cosw  +  (7,    IV.    I  cosec2^-  —  =  —  cotw-f-O, 
J  dx  J  dx 

V.    (  tan  w  •  sec  u  •  —  =  sec  u  -f  C, 
J  dx 

VI.    I  cot  u  •  cosec  u  •  —  =  —  cosec  u  -f  C, 
J  dx 

VII.    I  sin  u  •  -—  as  vers  ^*  -f  C, 
J  da; 

VIII.    f cos  tc .  —  =  -  covers  u  +  C. 
J  dx 

Example  I.  —  Integrate  ^  =  x  •  sin  (2  a;2), 
da; 

This   derivative  has   the   general   form  smu — .     Placing 

du  dx 

u  =  2x2,  —  =  4  x,  and  the  given  derivative  may  be  written 

(XX 


108       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


-£  =  J .  sin  (2  cc2)  •  4  x.     Hence,  by  the  formula, 
cix 


/- 


sin  w  •  —  =  —  cos  u  -h  C,  y  =  —  \  cos  (2  cc2)  +  O. 
dec 

dy        •   5 

•~  —  cm0' 


Example  II.  —  Integrate  —  =  sin0  x  •  cos  x. 

dx 

Since  —sin x = cos cc,  —  =  sin5cc — since,  and  ?/=isinccc-f  (7. 
dx  dx  dx 


lit 

PROBLEMS 

egrate, 

1. 

^=cos(4cc). 
dx 

7. 

dy      1         1 

-T  =  _.cos-. 
aa;     or        x 

2. 

^=cos(2cc-5). 
dx           v            J 

8. 

^  =  3sin(5ec-7). 

3. 

^=^.cos(^  +  2). 
dx 

9. 

dy       •   o 

—  =  snree  •  coscc. 

dec 

4. 

^/=sec2(3cc  +  5). 
da; 

10. 

dy       .   i 

-^  =  S1I12  X  •  COS  X. 

dx 

5. 

£-*©• 

11. 

dy           o 

—  =  cos^ec  •  since. 

dec 

6. 

^=  cos  (fee),  cot  (fee). 
dx           V2    y         V2    ' 

12. 

-M  =  tan5  ec  •  sec2  x. 
dx 

Frequently  an  expression  containing  circular  functions,  when 
not  directly  integrable,  may  be  transformed  into  an  expression 
which  can  be  integrated  by  means  of  the  trigonometric  rela- 
tions sin2ec  +  cos2ee  =  1,  sec2ec  =  1  +  tan2ec,  sin2cc  =  %  —  i  cos  (2  ec), 
cos2ec  =  |  +  |  cos  (2  cc),  sin  ec  cos  x  —  ±  sin  (2  ec). 

The  last  three  relations  are  special  cases  of  the  formulas 
sin  cc  sin  y  =  -J  cos  (x  —  y)—  \  cos  (x  +  y), 
cos  x  cos  y  —  \  cos  (x  —  y)+\  cos  (ec  -f-  ?/), 
sin  x  cos  2/  =  \  sin  (cc  +  ?/)  +  -|  sin  (cc  —  y), 
which  are  frequently  useful. 


CIRCULAR  FUNCTIONS  109 

Example  I.  —  Integrate  -&  =  sin5  a;  •  cos2  a;. 
dx 

Write  —  =  sin5  x  •  cos2  x  =  sin4  x  •  cos2  x  •  sin  x 

dx 

=  —(1  —  COS2#)2-COS2X  —  cos  a 
dx 

=  —  cos2  a?— coscc-f  2  cos4  a;  —  cos  a;  —  cos8 a;  — cos  a;. 
dx  dx  dx 

Integrating  term  by  term,  y  —  —  i  cos3  a;  +  -f  cos5  a;  —  \  cos9  a;  +C. 

Example  II.  —  Integrate  -^  =  tan4  a;. 

dx 

Write  —  =  tan4  x  =  tan2  x  (sec2  x  —  1) 
dx 

=  tan2  a;  •  sec2  a;  —  tan2  a;  =  tan2a;  •  sec2  a;  —  sec2  a;  +  1. 
Integrating  term  by  term,  y  =  |  tan3  x  —  tan  x  +  x  +  C. 

Example  III.  —  Integrate  -^  =  sin4a;. 

dx 

Write  ^  =  sin4aj  =  Ji-icos(2aj)}2 
ctx 

=  J  -  i  cos  (2  a?)  +  J  cos2  (2  a?) 

=  J  -  |cos(2a;)  +  i  {*  +  £cos(4ar)} 

=  t ~~  i  cos  (^  *)  +  i  cos  (^  a;)- 
Integrating  term  by  term,  y  =  f  x  —  £  sin  (2  a;)  +  ^  cos  (4  a;)  +  (7. 

Example  IV.  —  Integrate  *fe  =  !***£. 

da;     cos4  a; 

Write  ^  =  ^H^  =  ?H^.-V  =  tan2aj.sec2a;.    Integrating, 
da;     cos4  a;     cos2  a;  cos2  a; 

y  =  Jtan3a;H-C 


110       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Example  V.  —  Integrate  -^  =  sin2  x  •  cos2  x. 

dx 

"Write  -^  =  sin2  x  •  cos2  x  =  1  sin2  (2  #) 
dx 

=  i  I  i  -  i  cos  (4  *) !  =  i  - 1 cos  (4  ?)« 

Integrating,  ?/  =  -|-  a;  —  ^  sin  (4  x)  +  C. 

Example  VI.  —  Integrate  -£  =  sin  x  cos  4  x. 

dx 

Write  -^  =  sin  x  cos  4  a;  =  i  sin  5  96  —  4-  sin  3  a;.      Integrating, 
dx 

y=  —  y1^  cos  5  a?  +  |-  cos  3  x  +  0. 


PROBLEMS 
Integrate, 

1.   ^  =  sin2  (3  x).  cos3  (3  x).        8.   ^  =  sin3  x  •  cos3  a;. 


2.    ^  =  sin3(ia;).  9.    ^  =  sin4 

dx  V2    '  do; 


jura- cos4  #. 


3.  ^  =  cos2  (2  a).  10.   ^  =  tan5a. 
dx  dx 

4.  ^  =  tan3  (2  a?).  11.   &*-  cos5  (3  x). 
dx  dx 

5.  ^  =  3cos3a;.  12.   ^  =  3a>sin2(a^ -cos3(a;2). 
dx  dx 

6.  ^  =  5sin2(£a;).  13.   &  =  a;2  ■  sec2  (3  x3). 
(Xaj  otaj 

7.  ^  =  sm'x.cos'x.  14.   *y  =  sJH*l. 
dx  dx     cos2  a; 

Show  that,  m  and  ?i  being  positive  integers, 

X2tt  /»2ff 

sin  (mx)  *dx  =  0.  16.     I      cos  (mx)  >dx  =  0. 


20 


CIRCULAR   FUNCTIONS  111 

X2rr 
sin  (mx)  •  sin  (nx)  -dx  =  0,  m  =£  n. 

X2;r 
cos  (mx)  •  cos  (n#)  •  dx  =  0,  m  =£  n. 

X27r 
cos  (rax)  •  sin  (nx)  •  dx  =  0. 

J^2tt  /»2n- 

cos2  (mx)  -dx  =  iT.  21 .     I      sin2  (mx)  •  dx  =  -jr. 

0  »y0 


Art.  39.  —  Integration  by  Trigonometric  Substitution 


First  derivatives  with,  respect  to  x  involving  V  a2  —  x2  may 
be  transformed  into  trigonometric  derivatives  by  the  substitu- 
tion x  =  a  •  cos  0 ;  those  involving  Va2  +  x2  by  the  substitution 
x  =  a  •  tan  0 ;  those  involving  Vx2  —  a2  by  the  substitution 
a  =  a  •  sec  6 ;  those  involving  V2  ax  —  or*  by  the  substitution 
x  =  a  (1  —  cos  0). 

dv  1 


Example.  —  Integrate 


da;     aj^-a*)* 


Placing  a;  =  a  •  sec  0,  —  =  a  •  sec 0  •  tan  0,  (x2—  a2)?  =  a  •  tan  0. 
d0 

Hence,  ^/  =  4/ .  d*      1.  cos20  =  i{±  +  icos(2tf)J. 

'  dd     dx    d6     a*  a3*2      2        v     " 

Integrating,  y  =  —  J 1 0  -f  -^  sin  (2  0)  j  -f-  0.     From  a;  =  a  •  sec  0, 

0  =  sec"1-,  sin (20)  =  2  sin0  cos  0  =  2-Yl  -  ^  =  ^4(x2-a2)i- 
a  x\       x2J        x2 

Finally,  J/  =  -L  .  sec"1*  +  -^(x2  -  a2f  +  C. 
2  a?  a     2a2x2 


PROBLEMS 

Integrate, 

d.V___! 9    dy_ 


1.    Z2.-: : 2. 


dx     (a2-x*)%  dx     x2(l  +  x2)* 


112       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Q     dy  =       x5 


dy  _ 

1 

dx 
dy_ 

a?( 

x2-l)}* 

dx 

(1 

3 

dx 


(1  -  x2)^ 


7.  dv =    g?    . 

8.  ^i=—^—. 

dx      (2  +  ar2)*  '    tfa      Vl+¥2 

9.    ^=^(l-^)i 
dx 

10.   lfe=r-  -  *       •     Substitute  x  =  a-  tan 6. 
dx     (a?  +  ar)2 

u.  gg=     !     . 

<to      (1  -  ar2)2 


12.  |^  =  ^Lz^.     Substitute  »*  =  sin  ft 

13.  Find  the  area  of  the  circle  x2  -f-  y2  —  r2. 

A  =  2  I  (r2  —  x2)^  •  dx.  Substituting  cc  =  r  •  cos  0,  and 
noticing  that  when  x  runs  through  all  values  from  +  r  to  —  r, 
0  runs  from  +  £  to  —  jj,  there  results 


cos20.d0  =  7r.rJ. 


14.  Find  the  area  of  the  ellipse  —a-\-^-—l. 

a2     b2 

15.  Find  the  area  of  the  hypocycloid  x$  +  y$  =  a* 

16.  Find  the  volume  of  the  solid  generated  by  the  revolu- 

8  a3 

tion  of  v  = =  about  the  X-axis. 

9      x2  +  4a2 

x* 

17.  Find  the  area  bounded  by  the  curve  y2  = and  the 

3  2  a  — x 

line  x  —  2a.    A  =  I      -•     Substitute  x  =  2  a  •  sin2  0. 


Jr-*2a 
0 


(2  (*-•»)* 


CIRCULAR  FUNCTIONS 


113 


Art.  40. — Polar  Curves 


Tangents  and  Normals.  —  Let  r  =  f(0)  be  the  equation  of  a 
continuous  plane  curve,  P(r,  0)  any  point  in  the  curve,  and 
P'(r-\-Ar,  9  +  A0)  any  other  point  in  the  curve.  The  ratio 
measures  the  average  rate  of  change  of  r  in  the  interval 


Ar 
A0 


A0,  and  —  =  limit  —  when  limit  A0  =  0  measures  the  actual 
9  d$  A0 

rate  of  change  of  r  at  the  point  (r,  0). 

The  secant  through  (r,  0), 
(r  +  A?*,  0  +  A0)  approaches 
the  tangent  to  the  curve  at 
(?',  6)  as  A0  approaches  zero. 
Hence  the  angle  AP'S  ap- 
proaches the  angle  APT  =  <£ 
included  by  the  tangent  at 
(r,  0),  and  the  radius  vector 
to  the  point  (r,  0)  when  A0 
approaches  zero.  Drawing  a 
perpendicular  from  (r,  0)  to 
AP', 


tan  <j>  =  limit  tan  APrS  =  limit 


=  limit 


r  •  sin  A0 


r  +  Ar  —  r  •  cos  A0 


sin  A0 
A0 


A0 
rsm— - 

2     .    A0  .  Ar 


— X 


—  r 


eZ0 
dr 


when  limit  A0  =  0. 


114       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  distance  AT  from  the  pole  to  the  tangent  measured  on 
the  perpendicular  to  the  radius  vector  to  the  point  of  tangency 

is  called  the  polar  subtangent ;  the 
distance  AN  from  the  pole  to  the 
normal  measured  on  the  same  per- 
pendicular is  called  the  polar  sub- 
normal.   From  the  figure  subnormal 


AN=—;  normal  PN 

=  (W 

dr*\l 

dp)'' 

subtangent 

AT-. 

=  r> 

dO 
._;    perpen- 

dicular  from  poh 

j  to 

tangent 

AD  - 

=  p  = 

r2 

7*  + 

vie2) 

Asymptotes.  —  When  in  a  polar  curve  r=f(0),  for  some 
finite  value  of  6  =  00,  r  becomes  infinite  and  the  subtangent  is 
finite,  the  tangent  to  the  curve  at  infinity  passes  at  a  finite 
distance  from  the  pole  and  is  called  an  asymptote.  If  the  sub- 
tangent is  positive,  lay  it  off  to  the  right  of  the  radius  vector 
looking  towards  the  infinitely  distant  point  of  the  curve ;  if  neg- 
ative, lay  off  the  subtangent  to  the  left  of  this  radius  vector. 


Example.  —  Examine  r  —  -  for  asymptotes. 


For  0  =  0,  r  =  oo. 


The  subtangent  =  i* .  ^  = 
dr 


a.     Hence 


for  6  =  0,  subtangent  =  —a,  and 
===  the   asymptote   is    obtained    by 

laying  off  on  the  perpendicular 
—  to  the  polar  axis  at  the  origin  to 

the  left  when  facing  in  the  direc- 
tion $  =  0  the  distance  a  and  drawing  the  perpendicular  PT. 


Fig.  89. 


CIRCULAR  FUNCTIONS  115 


PROBLEMS 


1.  Find  the  subtangent  of  r  =  a-$. 

2.  Find  the  subtangent  of  r2  =  a2  •  cos  (2  0). 

3.  Examine  r  =  —  for  asymptotes. 

4.  Find    the    angle   under   which   the   radius   vector  cuts 

r=       P      - 
1  —  cos  0 

5.  Show  that  the  radius  vector  cuts  r  =  ae  under  a  constant 
angle. 

Length.  —  Denoting  the  length  of  the  curve  r=f(ff)  from 

0  =  $0  to  6  =  01}  by  8 


Fig.  40. 


0  =  0i      

=  limit  S    Vr2  •  sin2  A0  +  (r  +  Ar  -  r  •  cos  A0)2 

0  —  0o 

-  limit's*  JrfmM)%+  I'd- «* **)  ±±L\2Ae 
0=00  *    \   &0   J      \  A0  J 

=  f  *  \/^  +  ^= •  ^j  when limit  A#  =  o. 

Je=e0    *        d$2 


dO 


116       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Area.  —  Denoting  the  area  bounded  by  the  curve  r  =/(#) 
and  the  radii  vectores  to  the  points  (r0,  00),  (fa  0j)  by  A, 

A  =  limit  i  1 i*. A$  =  4  f  r*.  d0. 

0  =  00  c/00 

Example.  —  Find  the  length  and  the  area  of  the  cardioid 
r  =  a  (1  +  cos  6). 

The  length 
s  =  2  f  Yr2  +  ^ |V . d0  =  2  f* |  (a  +  a-  cos 0)2  +  a2- sin2<9  Ud0 

=  2a  f'rV2(l+cos^).^=4a-cosi^.d^=8a(sini^=8a. 
The  area  A  =  2 .  \  f  V  •  tf0  =  a2  C*  (1  +  cos  0)5 
=  a2  r*(l  +  2cos<9  +  cos20)  -d0 
=  a2  f7r(§  +  2cos0  +  icos20)d0 
=a2|fl9  +  2sin0  +  ^sin20r  =  f73 


PROBLEMS 

1.  Find  the  area  of  the  lemniscate  r2  =  a2  •  cos  2  0. 

2.  Find  the  area  of  r  =  2  a  •  sin  0. 

3.  Find  the  length  of  r  =  a  •  sin3-. 

4.  Find  the  area  of  one  loop  of  r  =  a  •  sin  (2  0). 

5.  Find  the  area  of  r  =  a  •  sec2  -  from  0  =  0  to  0  =  -. 

L  1; 


CIIiCULAR  FUNCTIONS 


117 


Art.  41.  —  Volume  of  a  Solid  by  Polar  Space 
Coordinates 

The  polar  space  coordinates  of  a  point  are  r,  <j>,  and  6.  The 
conical  surfaces  corresponding  to  6  and  0  +  A0  include  a 
conical  wedge  of  the  volume 
to  be  determined.  The  planes 
corresponding  to  <£  and  <f>  -\-  /\cj> 
cut  from  this  wedge  a  solid  of 
the  nature  of  a  pyramid.  The 
spherical  surfaces  corresponding 
to  r  and  r  +  Ar  cut  from  this 
pyramid  an  element  of  solid 
which,  when  A?*,  A<j>,  and  A0 
are  indefinitely  decreased,  ap- 
proaches as  its  limit  the  rec- 
tangular parallelopiped  whose 
ad  =  r  •  cos 6 •  A<£,  ac=  r  •  A0.  Hence  in  the  limit  the  element 
of  volume      =  r2  •  cos  6  •  dr  •  d<j>  •  dO, 

the  pyramid  =1        r2  •  cos  9  •  dr  •  d<f>  •  d0, 

•/r  =  0 

the  wedge      =1  I        r2  •  cos  0  •  dr  -  d<f>  >  d$, 

J<})=0       Jr=0 

and  the  entire  volume 


Fig.  41. 


dimensions     are     ab  =  Ar, 


I  (        r2  •  cos  (9  •  dr .  dd>  •  dO. 

=  0      c/0=O       c/r  =  0 

For  example,  let  it  be  required  to  find  the  volume  of  the 
trirectangular  spherical  pyramid,  the  radius  of  the  sphere 
being  a.     In  this  problem  r  extends  from  0  to  a,  6  from  0  to  ^, 


<£  from  0  to  |. 


Hence, 


118       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 
V=  (      2  I       2  (        r».cos0.dr.(fy.d0 

3    /»0  =  -      /»<*>  =  - 

2cos  0-d<£-d0 


a3  p=f  ' 

3»/0=o    c/<^=o 

I .  d0  « 2 
6 


6  */e=o 


Art.  42.  —  Differentiation  of  Inverse  Circular 
Functions 

To  differentiate  y  =  sm-1w,  where  w  is  a  continuous  function 
of   jc,  write   sin  y  =  u,  and   differentiate  with   respect  to  x. 

du 

There  results   cos  y  •  2a  =  — ,   whence  —  = ;    and,  since 

dx     dx  dx     cosy' 

du 
du      d    .  _i  da; 


cosy  =  VI  —  sin2 y—  Vl  —  w2,  -^  =  — sin  *w 


da;     da;  y  i  _  u* 

du  du 

T    ,.,  d         %  dx  d. .  da; 

In  like  manner  — cos-1w  = ;     —-tan  lu 


dx™  VI  -u2'    dx                1+w2' 

du  du 

d      l  i                   *»  d          ,                  dx 

—  cot_1w     =  —  rr-. — 5;  —sec-1^ 


da;  1+u*'  dx  uVw'-l' 

_du  du 

d  ,  da;  d  1  da; 

—  cosec-1  u  = ,  ;         —  vers-1  u  = 


da;  tfcVw2-l  «f  V2w-w2 

_dw 

d  _i  da; 

covers  lu 


dx  V2w 


CIRCULAR  FUNCTIONS  119 


2  x 
Example.  —  Differentiate  tan  x 


1-x2 


.      o  (l-x2).  —  (2x)-2x~  (1-x2) 

d     2x  v         J  dxK     }  dxK         ) 

d.     -i_2x__       dxl-x2  (1-x2)2 


tan 


dx  1—x2     -.       /   2  x  V  ^    ,       4^ 


2a?  V 

l-a?V  "  '  (1-a2)2 

=  2(l-a^)+4a;2=    2(1 +ft2)  2 

(l-^)2+4«2       l+2x2  +  x*     1  +  x2' 

d  2  x  d 

It  appears  that — tan-1 -  =  2 — tan_1#,  which   is  as  it 

dx  1  —  x2        dx 

2x 
ought  to  be,  since  tan  *  =  2  tan"1  x  by  trigonometry. 

J.  —  X> 


)ifferentiate, 

PROBLEMS 

1.    sin"1  (3  a;). 

A         4-             I2® 

4.   tan  1— • 
o 

7.    sec-1  (a;2). 

2.   3  sin-1??. 

5.   5vers-1-- 
5 

8.    COt-g). 

3.    sin"1  (3 x  +5). 

6.   3^2  +  2  cos-1  x. 
10.   cot-1^-^. 

9.   5  tan-1— 

X 

1 1 .  Show  that  —  sin-1  (3  z  -  4  Xs)  =  3  —  sin"1  x. 

dx  dx 

12.  Form  -^  and  — |  of  equation  of  cycloid 

cix  dx 

x  =  r  vers-1-  —  V2  ry—y2. 
r 

Here       ^  =  j        ,  henee  &  =  J?H^. 

<*S»        ^/2ry-y'  dx       v     y 

Squaring,   fg-  ?T- 1.     Differentiating,  2  &  g  =  -  ?T  & 
dx2       y  dx  dx2  y2  dx 

Whence,     ^1  =  -L. 
dx2         y2 


120       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


13.    Form 

—  and  —0  of  cycloid  y  =  r>  vers-1-  -f-  V2  rx  —  x2- 
dx  dx2  r 


14.   Find  length  of  cycloid  y  =  r  •  vers-1-  +  V2  rx  —  x2. 


x 
r 

15.  On  a  pedestal  25  feet  high  stands  a  statue  11  feet  high. 
Find  the  distance  from  the  base  of  the  pedestal  of  the  point  in 
the  horizontal  plane  through  the  base  at  which  the  statue 
subtends  the  greatest  angle. 

Art.  43.  —  Integration  by  Inverse  Circular  Functions 

The  results  of  the  preceding  article  when  -  is  substituted 
for  u  may  be  written  in  the  form, 

du  du 

T      r      dx  .     ,u  ,    ~      TTT     f    dx         1         .ti 

I.     I      .  =  sin"1-  +  C-,     III.     I  -5- — 5  =  -tan-1-  +  C: 

J  ^/a2  —  u2  a  J  a2  +  u2     a  a 

_du  _du 

II.     I     ;  =  cos-1-  +  C\     IV.     1-2- — s  =  -cot-1-+C; 

J  Va2  -  u2  <*>  J  a2  +  u2     a         a        ' 

V.     I  — ;  =  -sec-1-  -f-  C; 

<2tt 

VI.     I —  =  -cosec~1-  +  C; 

J  u-Vu2  -  a2     a  « 

VII.     f    -    dx         =  vers-1- +(7; 
J  V2  aw  -  u2  a 

du 

VIII.     f    -     da;       =  covers-1- +  a 
J  V2au-u2  a 


CIRCULAR  FUNCTIONS  121 

Example.  —  Integrate  -^  =  .     This  derivative  has 

dx     V2  -  4  ar3 
du 

the  general  form  — —  Placing   w2  =  4a^,   w  =  2a;*  and 

V  a2  —  u2  j        . 

f!  =  a**.     Writing  ^  =  —^  =  i-^=> 
c?a?  &d«     V2-4JB8     3V2-4«* 

?/ = -^  sin"1  (V2-a;f)  +  C. 


PROBLEMS 

Integrate, 

dy_       3  % 


cto     4  +  9  a;2  da;     1  +  a4 

2.    ^  =  __J__.  4.    ^=            3 

*»     a;V3a;2-5'  "   *c     V5a;4-2a/ 

d?/  x 

5 .   -£  = — .     Reduce  improper  fraction  to  mixed  number. 

cix     1.  -j—  xr 

6     ^  —      x  a    dy  _  tan-1  a; 

da;      a?4  +  4  '   dx      1  +  x2 


7.    <fy=         1  10#    ty  = 


sin-1  a; 


dx     VGaj-ic2  *»      Vl-ar8 

o     #_        a;  dy _    sec-1  x 

dx      Vl  —  x4  '    d®     a^Var2  — 1 

—  (a;-3) 

12     <*?  =  1  Since  ^/=__J_  =  i^ , 

da;     ^-6^  +  11  da;     2  +  (a;-3)2     2+(a;-3)2 

2/  =  ^tan-1^  +  0. 
V2  V2 

X3.   ^/__      _i_     _. 


*B      Vl+3a;-a^ 


122 


DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


14.    The  time  of  descent  of  a  body  down  the  arc  of  the 
vertical  frictionless  cycloid  x  =  r  •  vers-1^  -f-  V2  ry  —  y\  from 


Fie.  42. 


y  =  h  to  the  vertex,  is  t  =  ( -  J2  (        .  Show  that  the 

time  is  the  same  for  all  positions  of  the  starting  point. 

15.  A  body  is  suspended  on  a  frictionless  horizonal  axis, 
turned  through  a  small  angle  0O,  and  then  left  free  under  the 
action  of  gravity.  This  constitutes  a 
compound  pendulum.  The  relation  be- 
tween 0,  the  angle  the  pendulum  in  any 
position  makes  with  the  vertical  and 
the  time  t  measured  in  seconds  after  the 
pendulum  is   started,   is   expressed  by 


the  equation 


ah 


0,  where  g  is 


Fig.  43. 


complete   oscillation. 

dO 

dt 


dtf  h*  +  kf 
the  acceleration  of  gravity,  h  and  kx  are 
constants  depending  on  the  shape  and 
material  of  the  pendulum  and  the  posi- 
tion of  the  axis.  Find  the  time  of  a 
In  this  problem,  when  t  =  0,   6  =  00, 


0. 


CIRCULAR  FUNCTIONS 


123 


16.  In  strength  of  materials  it  is 
proved  that  for  any  point  (x,  y)  of  the 
elastic  curve  of  a  long  column 


EI 


fy_. 

dx2 


-P-y, 


where  E  and  /  are  constants  depending 
on  the  material  and  cross-section  of 
the  column;  P  is  the  load.  Calling 
the  maximum  deflection  A,  when  y  =  A, 

-^  =  0,  and  when  x  =  0,  y  =  0.  Find 
dx 

the  equation  of  the  elastic  curve. 


Art.  44.  —  Eadius  of  Curvature 

If  a  point  moves  in  the  circumference  of  a  circle,  the 
tangent  to  the  circle  at  this  point  changes  its  direction.  Sup- 
pose the  point  to  start  from  A.  When  it  reaches  B,  the  tangent 
has     turned     through     the     angle  . 

T'ST=AOB.  The  ratio  of  the 
angle  AOB  to  the  distance  AB  the 
point  has  moved  along  the  circle  is 
called  the  rate  of  curvature  of  the 

6       1 
circle,  and  equals  — -  =  -•    That  is, 

9  *        r-B     r 

the  rate  of  curvature  of  the  circle 

is  the  reciprocal  of  the  radius  of 

the  circle. 

If  a  point  moves   along  any  curve  y=f(x),  the  ratio  of 

the  angle  through  which  the  tangent  turns  to  the  distance 

the  point  moves  is  called  the  average  rate  of  curvature  of  the 

curve  for  the  distance  the  point  moves.      Denoting  by  A<£ 


Fig.  45. 


124       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

the  angle  through  which  the  tangent  turns  while  the  point 
moves  a  distance  As  along  the  curve,  the  average  rate  of  cur- 
vature is  — &     The  actual  rate  of  curvature  at  any  point  {x,  y) 

of  the  curve  is  the  value  at  (x,  y)  of  limit  — *  =  --2  when 
limit  As  =  0. 

The  reciprocal  of  the  rate  of  curvature  at  any  point  (x,  y)  of 
the  curve  is  the  radius  of  the  circle  which  has  the  same  rate 
of  curvature  as  the  curve  at  the  point  (x,  y).  The  reciprocal 
of  the  rate  of  curvature  is  called  the  radius  of  curvature,  and 

is  denoted  by  p,  so  that  p  =  — .     Now  <£  =  tan-1-^,  hence 

d<f>  dx 


ds 

ds 

dx 

ds 
dx 

(»+» 

(■+£)' 

p 

dcj>~ 

dx 

d<f> 

~d$  = 
dx 

d2y        - 
dx2 

dx2 

d2y 
dx2 

The  analysis  supposes  that  the  curve  is  such  that  y  is  a 
continuous  function  of  x,  and  <f>  a  continuous  function  of  s. 
If  the  equation  of  the  curve  is  given  in  the  form  x=f1(t), 
dy 


"*»  1= 

.*    and 

dx 

dt 

d2y 

d  dt 

dy             d2y  t  dx     d2x  t  dy 
d  dt      1       dt2  '  dt      dt2  '  dt 

dx2 

' dxdx~ 
dt 

dt  dx    dx~              dx3 
dt     dt                  d? 

Hence 

P  = 

(dx*     df\\ 
[dt2  T dt2) 

dt2  "di      df  "dt 

CIRCULAR  FUNCTIONS 


125 


For  a  polar  curve  r  =/(0),  x=r-cos$,  y  =  v sin 0.    Hence 
|=-,.sin,  +  cos,.|, 

§=,.cos*+sin*.|, 


tf02 


=  —  r-sin0  +  2sin0 


d0^  d02 


Hence  for  a  polar  curve  p  = 


T    d02  d02 


A  circle  of  radius  p,  placed  so  that  the  circle  and  curve 
V  —f(x)  have  a  common  tangent  at  (x,  y)  and  lie  on  the  same 
side  of  the  common  tangent,  is  called  the  circle  of  curva- 
ture of   y  —  f(x)  at   (x,  y)  ; 
the  center  of   the  circle   is 
called  the  center  of  curva- 
ture  of  y=f(x)   at  (x,  y). 
Denoting  the  coordinates  of 
the  center  of  curvature   by 
a  and  /?, 
(1)  a  —  x  —  p  •  sin  <f> 


dx\       dx2)  t 


da2 


(2)  0  =  2/  +  p  '  cos  <£  =  y  + 


126        DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  locus  of  the  center  of  curvature  as  the  point  (x,  y)  traces 
the  curve  y=f(x)  is  called  the  evolute  of  y  =  f(x).  The 
equation  of  the  evolute  is  found  by  solving  (1)  and  (2)  for  x 
and  y  in  terms  of  a  and  /3  and  substituting  in  y  =  f(x). 

Example  I.  —  Find  the  radius  of  curvature,  coordinates  of 
center  of  curvature,  and  evolute  of  the  parabola  y2  =  2px. 

Heref^,    f*  =  -&  hence  „  =  - i^+f^ ■  a  =  3x+p, 
ax      y     ax*         y*  pz 

(3  =  —  ^ ;  the  evolute  is  ft2  =  ^T  (a  —pf,  a  semi-cubic  parabola. 

The  radius  of  curvature  to  the  parabola  at  the  vertex  (0,  0) 
is  p;  at  the  extremity  of  the  latus  rectum  (^p,  p)  the  radius 
of  curvature  is  2  V2  -p. 

Example  II.  —  Find  the  radius  of  curvature  of  xy  —  4  at 
the  point  (1,  4). 

For  this  curve  ft —  4,    ^=  1     At  (1,  4),  ft=  -4, 

dx  x2     dx2      x3  dx 

g=8.    Hence  P  =  i  (2)1. 

PROBLEMS 
Find  the  radius  of  curvature  of 

1.  jf  =  4*at(l,2);  (0,0);  (4,4). 

2.  •|  +  l2  =  lat(3,0);  (0,2). 

a2     62  a2     62 

5.  x  =  r  •  vers-1 V2  r?/  —  ?/2. 

6.  x  =  t>.*,  y  =  %g>l2. 

7.  x  =  acos<j>,  y  =  b sin $. 

8.  x  =  R'0-R-$me,y  =  R-K-cos$. 


CIRCULAR  FUNCTIONS  127 

9.    2  xy  =  a2.  12.    r  =  a  (1  —  cos  0). 

10.  <e*  +  y*  =  ai  13.    r2  =  a2  •  cos  (2  0). 

11.  ?/  =  sin  #.  14.    r  =  a0. 


CHAPTER  VIII 

LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS 

Art.  45.  —  The  Limit  of  1 1  -J-  -  J  When  Limit   z  =  cc 

Assume  first  that  z  takes  only  positive  integral  values  m. 
By  the  binomial  formula, 

\       mj  m  1-2        m2 

.  m  (m  —  1)  (m  —  2)  1_  • 

1-2-3  m3  +  '" 

.  m (m  —  1)  (m  —  2)  ■  •  »  (m  —  n  +  1)  1 

.  m  (m  —  1)  (m  —  2)  >  ♦  •  (m  —  n  + 1)  (m  —  ?i)    1 
l-2.3.4.»n.(n  +  l)  mn+1 

,  m(m  —  l)(m  —  2)  ...  jm  —  (m  —  l)j  J_ 
1  •  2  •  3  •  4  •  •  •  ra  mm 

=  1-1  .  m      V        m)\       mj 

1.2-3  + 


+KX 


m 


1-2. 3-  ..7i 
128 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS     129 
\      mj\      mj     \         m   J\       m     \      mj\         in  J 


+  1-2.3-n  U+l  +      (7i  +  l)(n+2)      + 

V        mj\  m   J      \  m 

+  (n  +  l)(n  +  2)-..m  J 

This  expansion  is  true  for  positive  integral  values  of  m  how- 
ever large  m  may  be  taken.  Denote  by  S  the  sum  of  the  first 
Ti  +  1  terms  of  the  expansion,  by  R  the  sum  of  the  remaining 
terms.     Then  the  equation 


limit  [  1  +  -  J  =  limit  S  +  limit  R 


K)' 


is  always  true.     When  m  is  indefinitely  increased, 
limit^  =  l  +  l  +  -J-+^4-^+,   }n   T  + 


1-2     1-2-3      1-2. 3-4  1.2- 3- 4- ..n 


+  (n  +  1)"  r 
when   m    is    indefinitely   increased,   limit  R  < 


n(1.2-3...n) 

Now  n  may  be  taken  so  large  that  limit  R  becomes  less  than 
any  quantity  that  can  be  assigned.     Hence,  when  limit  m  =  oo, 

lir»it(l+I)™ 

=  limitj1  +  1+_^  +  _T|_+i_-l_  +  ...i23l4  —  } 

when  limit  n  =  oo. 

K 


130       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  value  of  limit  f  1  -\ — ]    when  limit   m  =  oo    to  nine 

decimals  is  2.718281828.    This  limit  is  the  base  of  the  Napier- 
ian system  of  logarithms  and  is  denoted  by  e. 

Next  assume  that  z  may  take  fractional  as  well  as  integral 
positive  values  as  it  approaches  infinity.  Denote  any  one  value 
of  z  by  s,  and  let  m  be  an  integer  such  that  m  <  s  <[  m  +  1. 

Then  (1+riT)"<(1+«),<(1+y1'  whi°h  may  be 

/  1       \m+l 

„1„„t^_<(1+ij<(1+i)-.(l+i). 

ra  + 1 

When  limit  s  =  go,  limit  m  =  oo.     Hence  when  limit  s  =  oo, 
limit  fl-h-]     lies  between   two   quantities   whose   common 


limit  is  e.    Consequently  limit  (l  +  - j  =  e  when  limit  s  =  oo. 
Lastly  assume  that  z  —  —r,  and  that  limit   r=oo.     Now 

limit  (l-^^limit^-^limit^J 

=  limit(l+-AIJ-1.(l+-ly)  =  e 
when  limit  r  =  oo. 

It  appears  that  in  every  case,  when  limit  z  =  oo,  limit 
(l  +  -Y  =  e  =  2.718281828.  This  limit  is  fundamental  in  this 
chapter. 


LOGARITHMIC  AND  EXPONENTIAL   FUNCTIONS     131 


Art.  46.  —  Differentiation  of  Logarithmic  Functions 

Let  u  represent  a  continuous  function  of  x,  and  denote  by 
Au  and  Ax  corresponding  changes  in  u  and  x.  Then,  if  a  is 
the  base  of  the  system  of  logarithms  used, 

Afc>g.u  -  Hmit  log.("  +  Au)-log.u  .  *u 

dx  Au  Ax 


log, 


u  4-  Au 


limit 


Au 


Au 
Ax 


=  limit  -•log0[  1 
u  \ 


=  limit  -  •  loga 
u 


1  + 


U 

Au 


Ax 

a^    Au 
Au 


logae 


du 
dx 


when  limit  Ax  =  0. 


For  by  the  nature  of  logarithms  the  difference  of  the  loga- 
rithms of  two  numbers  is  the  logarithm  of  the  quotient  of  the 
numbers,  and  the  logarithm  of  a  number  affected  by  an  ex- 
ponent is  the  exponent  times  the  logarithm  of  the  number. 
Since   u  is  a  continuous  function  of  xf  limit  Au  =  0  when 

limit  Ax  =  0.   Writing  —  =  z,  when  limit  Au  =  0,  limit  z  —  oo. 

Au 


Hence  limit 


1  + 


u 

Au 


when  limit  Ax  —  0  equals  limit  ( 1  -f-    ) 


when  limit  %  =  oo  • 

Calling,   as   is   customary,  logae  the  modulus  of   the  sys- 
tem of  logarithms  whose  base  is  a,  and  denoting  it  by  M, 


132       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

du 

d  dx 

—  loga«  =  M — ;  in  words,  the  derivative  of  the  logarithm  of 
dx  u 

a  function  of  x  is  the  modulus  of  the  system  of  logarithms 

times  the  derivative  of  the  function  divided  by  the  function. 

In  the  Napierian  system  of   logarithms  —  logeM=i-—  • 

dx  u   dx 

Unless  otherwise  specified,  the  Napierian  system  is  to  be  used. 

Example.  —  Differentiate  log  a/    ~  x. 

*1  +  a: 


^\l-x      1  +  s)      1- 


-1_ 

x2' 


PROBLEMS 


Differentiate, 

1.  logo2.  4.   log(a^).  7.   log  tan  a?. 

2.  log-.  5.   log2*.  8-    logVl-ar*. 

X  j  n 

3.  log (3* -5).  6.    log  sin*.  9'    logI^- 

10.   log  tan  i*.  mmi    . 

*  13.    log 


11.    1ogJl-cos«. 

\1  + 


12.    log 


-cosx  14.  iog(»  +  vrr5). 

cos  x  15.   log  (x  +  V*2  +  a2)- 

16.   log^  +  Va^-a2). 


Find  the  true  value  of, 

17.   J5£»     when  0.1  18.    *2££  when  x  =  l. 

(1  -  »)i  *  -  1 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS     133 

19.   SKI  when  x=cc.  22.  a;mlogna;  when  x  =  0. 

xn  ° 

20-  Si when  ■ = °-  23-  ^i "  .*§=  when  * = L 

_,     log  tan  (2 as)     -.              A  _.  log  sin  a;       ,               tt 

21.   — 2 i — *  when  a;=0.  24.  — ^ — — - —  whena;  =  -. 

log  tan x  {Tr—2xy  2 


Art.  47.  —  Integration  by  Logarithmic  Functions 

The  result  of  the  preceding  article  may  be  written 
du 


/"*-/-•£-■*•♦« 


Example.  —  Integrate  -* 


da;     1  -f-  x2 
The  first  derivative  of  the  denominator  is  2  a?,  hence 

—  (1  +  a,-2) 


PROBLEMS 

Integrate, 

1     dy_   l  +  3a;  dy  _  2  g  +  3 

da;     2  a;  +  3  a;2*  '   dx     x2  +  3x 

ft.   */=    3^_.  6.   *  =  tan*. 

da;     ar  +  7  da; 

3     dy^nxdx  6    ^  =  cotx 

da;     a2  +  a^  da; 


7.   &=  J_.     Writing  &  =  J-  = 


daj     sin  a;  da;     sin  x     2  sin  £  x  •  cos  J  a; 


d  /,  „x       d 


sec2^a; —  (±x)     — tan^a; 

.£"*!*  = _^ =  ^__)2/  =  logtan^+a 

tan^a;  tan^a;  tan^a; 


134 


DIFFERENTIAL   AND  INTEGRAL   CALCULUS 
dy_         1 


8.  dy  =  J_ 

dx     cos  x 


dx     x(l  +  x2)' 


Substitute  x  =  tan  6. 


10.   4  = 

dx     Va;2 


Write  Va2  +  a2  =  z  —  x,  whence 


/-n— — o  dx     z  —  x   dy     dy  dx 

z  =  x+-Vx2  +  a2,  —  = ,  -f-=-f-.  —  = 

dz         z       dz     dx  dz 

y  =  log.z  +  C=\ogM(x+Vrf  +  c?)  +  (7. 


1       z  —  #    '  1        a 

=  -,  and 

z  —  x      z        z 


11.   ^  =  — J 

*»      Va2  -  a2 


Writing 


12.    4- 


da      Vz2  +  2a 
d 


# 


C?# 


(•+1) 


Vz2  +  2a     V(>+1)2-1     <**     V(x'  +  1)2-1 


i  and 


y  =  log  {(a  + 1)  +  vV  +  2sj  +  C. 


13     ^.V  =  2  +  3^ 
'    da?      1  +  x2 
term  by  term. 

dy  _  m  -\-nx 


Write  ^  = 


dy_     2 


+ 


3* 


da;     1  +  a2     1  +  ar 


and  integrate 


14. 


17. 


dx      a2  +  x1 

dy  _     x3 
dx     x*  +  1 


dy  _  log2  a; 
dx         x 


16. 


dy _ logw  x 
dx        x 


Keduce 


& 


to  a  mixed  number. 


Fig.  47. 


a^  +  l 

18.    Find  the  area  of  xy  —  1  from  #  =  0  to 
x  =  x\ 

r       19.    If  A  is  the   cross-section  of  a  bar  of 
1    uniform   strength   at   a   distance   y  from   the 

(1  j\  7/i  .yd 

lower  end,  —  = ,  where  w  is  the  weight 

1  dy         8 

of  the  bar  per  cubic  foot,  and  S  is  a  constant 

depending  on  the  material  of  the  bar  and  the 

area  of  the   lower   end.      Find   the   relation 

between  A  and  y. 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS     185 


Art.  48.  —  Integration  by  Partial  Fractions 

F(x) 
The  function  of  x  denned  by  — j-f  is  called  a  rational  frac- 

tion  when  x  does  not  occur  affected  by  a  fractional  exponent 
or  under  a  radical  sign. 

If  the  numerator  of  the  rational  fraction  — ^2   is  not  of 

lower  degree  than  the  denominator,  the  given  fraction  may 
always  be  transformed  by  division  into  the  sum  of  an  integral 
function  F(x)  and  a  rational  fraction  J  K  J,  whose  numerator  is 
of  lower  degree  than  its  denominator.     For  example, 

a2  +  l~  a^  +  1 

f(x) 
Consider  the  rational  fraction  2-^2  whose  numerator  is  of 

.♦(*) 

lower  degree  in  x  than  the  denominator.  Suppose  the  denomi- 
nator <£  (x)  to  be  of  degree  n,  then  f(x)  cannot  be  of  higher 
degree  than  n  —  1.  By  a  theorem  in  algebra  <f>  (x)  can  be 
resolved  into  n  factors  of  the  first  degree,  and  imaginary  fac- 
tors must  occur  in  conjugate  pairs.  The  product  of  a  pair  of 
conjugate  imaginary  factors  of  the  first  degree, 

(x-c-d  V^l)  (x  -  c  +  d  V^l)  =  (x-  c)2  +  d2, 

a  real  factor  of  the  second  degree.  Hence  <£  (x)  can  be  resolved 
into  real  factors  of  the  first  and  second  degrees. 

Let  <£  (x)  =  (x  -  a)  (x  -  b)*\(x  -  cf  +  d*\\(x-e)*  +f2\'-     It 

is  proposed  to  break  up  the  fraction  21?  into  the  sum  of  par- 

tial  fractions  whose  denominators  are  the  factors  of  <f>  (x),  and 
in  every  partial  fraction  the  numerator  is  to  be  of  lower  degree 
than  the  denominator. 


136       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Assume 

f(x)__     A  B,  B2  B  Cx  +  D 

~<f>  (x)      x~a     (x-b)s      (x-  by-1  x  -  b      (x  -  c)2  +  d2 

E,x  +  Fx  E2x  +  F2  Etx  +  Ft         , 

\{x-  ef+f\<  "*"  l(x-ey+f2l<-^      (x  -  e2)  +  /2 

The  number  of  undetermined  constants, 

A,  Bl7  B2,  -",  Bs,  C,  D,  E1}  Flf  E2,  F2,  •  ••,  Et,  Ft 

introduced  equals  the  degree  of  cf>(x).  Multiplying  both 
members  of  the  identical  equation  by  <£(#),  the  right  hand 
member  will  be  of  degree  n  —  1  in  a;,  while,  by  hypothesis, 
f(x)  cannot  be  of  a  higher  degree  than  n  —  1.  Collecting  the 
terms  of  like  powers  of  x  in  the  right  hand  member,  the  co- 
efficients of  the  resulting  n  terms  are  linear  in  the  n  undeter- 
mined constants.  Hence,  by  equating  the  coefficients  of 
corresponding  terms  of  both  members  of  the  identity,  there 
result    n    equations    linear    in    the    n    assumed     constants. 

These  equations  determine  the  assumed  constants  uniquely. 

fix) 
Hence  the  fraction  J  \  {  can  be  broken  up  into  partial  frac- 
as) 

tions  of  the  form  assumed  and  in  only  one  way. 

A  few  examples  will  explain  the  process  of  breaking  up  a 
fraction  and  show  the  importance  of  partial  fractions  in 
integration. 


Example  I.  —  Integrate  -^  =  — 


-1 


dx     x2  —  4 

3j3 I  ic3  —  1  4,x  —  1 

Transforming  — to  a  mixed  number,  — =  x  +  — — -  - 

x2  —  4  x2  —  4  x2— 4 

A                     4a-l        A      .      B 
Assume  — = -f 


aj2  _  4      x  +  2      x-2' 
whence  ±x-l  =  (A  +  B)x+(-2A  +  2B). 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS      137 

Equating  the  coefficients  of  like  powers  of  x, 

A  +  B  =  2,    -2A  +  2B  =  -1,   and^  =  f,    B=\- 

ttq  nt%    dy     x>-l  .Ax-1  ,9     1       ,7     1 

Hence,  -£  =  — =  as  H — =  x-\ 

'  dx     ar>-4  a;2-4  4a>  +  2     4a>-2 

Integrating,  y  =  |  +  J  log  (x  +  2)  +  J  log  (a?  -  2)  +  G 

Example  II.  —  Integrate  ^  =   05a?  +  1    ■ 
6         da;     a^  +  a?  -  2 

Assume         5aJ  +  1  ^      ■      B 


x?  +  x-2      a;  -  1      a;  +  2' 
whence  5  x  +  1  =  A  (x  +  2)  +  B(x  —  1). 

This  identity  is  true  for  all  values  of  x.     When  x  =  1,  A—2\ 

when  a;  =  -  2,  £  =  3.     Hence,  ^  =  -^—  +  — —     Integrat- 
ed    x  —  1      a;  +  2 

ing,  y  =  2  log  (as  —  1)  +  3  log  (a?  +  2)  +  0.    This  result  may  also 
be  written, 

y  =  log  (a;  +  l)2  +  log  (a?  +  2f  +  log  c 

=  log{c(a;-l)2(a;  +  2)3|. 

Example  III.  —  Integrate  ^  =    a^  +  4a;-2 
5         dx     l+x  +  tf  +  x? 

Assume       /  +  4g-2        ^_  +  ^  +  C 
1  -f  a  +  a;  -  +  ar3      1  +  aj       1  +  x2 

whence  ar2  +  4  a;  -  2  =  (^  +  B)x2  +  (5  +  0)as  +  A. 

Equating    the   coefficients   of    like    powers   of    x,   A  =  —  2, 
£+0=4,    A  +  B  =  l,    whence   ^  =  -  2,    B  =  3,    0=1. 

Hence,  ^  =  ^!  +  ^L_  +  _L_; 

(to     l  +  a>     1  +  a;2     1  +  a^5' 

and       y  =  —  2  log  (1  +  a?)  +  f  log  (1  +  af)  +  tan"1  a;  +  C. 


138       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Example  IV.  —  Integrate  ^l  =  t± 


-1 


dx       (a2  +  2)2 
A  x*  +  x  —  l_Ax  +  B  |   Cx+  D 

Assume        (x*  +  2y  =  (^T2)"2 + ~^T2" 

By  clearing  of  fractions  and  equating  the  coefficients  of  like 
powers  of  x,  it  is  found  that  A  =  —  1,  B  =  —  1,  (7=1,  Z>  =  0. 

Hence    dy  =  ~x~1    i       g      =      ~x       i       g 1 

9  dx      (x2  +  2)2     ar>  +  2      (ar2  +  2)2      ar2  +  2      (ar2  +  2)2 

The  first  two  terms  are  directly  integrable.     To  integrate  the 
last  term  substitute  x  =  V2  •  tan  0. 


PROBLEMS 

Integrate, 


dx     x* 


2.    ^  = 


3.    ^ 


dy  _ 

2a  +  3 

dx 

x3  +  X2  —  2x 

dy  _ 
dx 

a 
x2—  a2 

dy  _ 

2-Sx2 

dx 

(x  +  2)2 

dx 

X2 

x4  +  x2-2 

dy  _ 

1 

dx 

x*-x2  +  2x-2 

dy  _ 
dx 

1 

Sho 

du 
wthat    C     dx    - 

dy  _      1 


dx     x  (1  +  x2) 
o     dy  _  x 


dx     x4  —  x2  —  2 


.  &= i 

da!     oj  (1  +  a;)2 
dy  _  3  x  + 1 


5.    ^  =  - ^— -•  11 


6.    2*  =    „         •  12. 


cfcc      a;4  —  1 

d?/_     a;4 
dx     x2  —  1 


.,  =  77-  log \-  G,  u  being  a  con- 

J  ft  —  or     2  a       u  +  a 

tinuous  function  of  x.     This  result  is  very  useful. 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS     139 

14.  Integrate ^=-=-| -•    Writing  f  =  ^\~     ., 

dx     x>-6x  +  5  dx     (x  —  Sy  —  i 

by  Prob.  13,  y  =  |  log  |*~g~*  +  C=  J  log  |^|  +  0. 

15.  Integrate  1  =  ^^. 

Art.  49.  —  Integration  by  Parts 

From   —  (u  •  v)  =  u  •  —  +  v  •  —   is   obtained  by   integration 
ctev       '  dx         dx  J  & 

Ju-  —  =  u*v  —  I  v-— ,  which is  called  the  formula  for  inte- 
rs; J       dx 

gration  by  parts.     The   following   examples   will   show  the 
application  of  this  formula. 

Example  I. — Integrate    -^  =  x-log#. 

dx 

Writing  u  =  log  x,  —  =  x,  whence  —  =  -,  v  —  A  a;2  -j-  Cl}  the 
dx  dx     x 

application  of  the  formula  gives 

y^fx.logx^tf+CJ-logx-fdxt  +  C,).! 

=  %x2.logx+C1'\ogx-%x2-C1-logx+C 
=  %x2-logx  —  {x2  +  C. 

CL  the  constant  of  the  integration   |  — ,  always  eliminates  as 

J  dx 

in  this  example.    It  may  therefore  be  neglected. 

Example  II.  —  Integrate  -^  =  a;  •  sin  x. 

dx 

di)  flu 

Writing  u  —  x,  —  =  sin  x,  whence  —  =  1   and  v  =  —  cos  x, 
dx  dx 

the  application  of  the  formula  gives 

y  =  J  a:  •  sin  a;  =  —  a;  •  cos  x  -f-  I  cos  x  =  —  x  •  cos  x  +  sin  x  -f  G. 


140       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Example  III.  —  Integrate  -^  =  x  •  sin-1  x. 

dx 

Writing     u=sm~1x,     —  =  x,     whence  —  = — — ===    and 
dx  dx     -y/J  _  x2 

v  —  \  x2,  the  application  of  the  formula  gives 

y  =  I  x  •  sin-1a;  =|-  a,2  •  sin-1  s  —  j-  I     -  • 

dlJ '  fly 

Write   —  =  and  substitute  a?  =  sin  0.     There  results 

ax      -y/i  _  #2 

%'  =  ^'  ^=  in2  ( 

dO      dx    dO  22V     /j 

and         y'  =  \0-\ sin (20)  +  C  =  \Q  -  isin0  •  cos0  +  C. 


Substituting  sin  0  =  x,  cos  0  =  Vl  —  x*,  0  =  sin-1  x, 


y'  =  l.  sin-1  a;  —  |-  a,  •  Vl  —  a2  +  (7, 


and  y  =  ^ar2  •  sin-1  a;  +  i  sin_1a;  —  i  a;  •  Vl  —  a2  -{-  C. 

Example  IV.  —  Integrate  -^  =  sec3a;. 

dx 

Writing  u  =  sec  x,  —  =  sec2 a,,  whence  —  =  sec  x  •  tan  a;  and 
da;  dec 

w  ss  tan  x,  the  application  of  the  formula  gives 

y  as  J  sec3  a;  =  sec  a;  •  tan  a;  —  I  sec  a;  •  tan2  a; 

=  sec  a?  •  tan  x  —  I  sec3  a;  -|-  j 

*/  •/  cos  a; 

Hence   2  |  sec3  a;  =  sec  a?  •  tan  a;  +  I  — — 
»/  J  cos  a? 

=  sec  a;  •  tan  x  —  log  tan  [  -  —  -  )  +  Q 
=  sec  x  •  tan  a;  +  log  (sec  x  —  tan  x). 


LOGARITHMIC  AND   EXPONENTIAL  FUNCTIONS     141 

fly  

Example  V.  — Integrate  ^=  Va2  +  x2. 


dv     -,         i  _     _   du 


Writing  u  =  Va2  +  x*.  —  =  1,    whence  — -  =  — ====  and 
dx  dx     Va2  +  x2 

v  =  x,  the  application  of  the  formula  gives 

Jx2 
VOLTS'" 

dy        a?  +  3?  a'      .  a? 

A  lso  — ==  —         —  =  —         —  -j-  —         — 

dx      Va2  +  a^      ^/a*  +  rf      Va2  +  x2, 

/x2 

Adding  (1)  and  (2)  and  solving  for  ?/, 


2/  =  -  Jaj.  Va2  +  ar*  +  ^  log  (a  +  Va2  +  a;2)  +  (7. 

PROBLEMS 

Integrate, 

i.   &=*.log*  5.   ^  =  121^. 

da;  da?        ^f 

2.  ^=o».cosa.  6.   ^  =  loga;. 
da;  dx        & 

3.  ^  =  tan-a;.  7.   ^  =  _^M^. 
da;  da;      ^  +  ^ 

.     dv  ,     2  o     dto     a?  tan-1  a; 

4.  -^  =  a;- tan2a;.  8.   -f-  =  — — . 

dx  dx        1  +  x2 

9.   Find  the  area  of  the  cycloid, 

x  =  r  •  6  —  r  •  sin  9,   y  =  r  —  r  •  cos  0. 

10.  Find  the  area  of  the  cjcloid, 

x  =  r  —  r  •  cos  0,   2/  =  *"  •  0  4-  J*  •  sin  0. 

11.  Find  the  volume  of  the  solid  generated  by  the  revolu- 
tion of  the  cycloid  x  =  r  •  6  —  r  •  sin  6,  y  =  r  —  r  •  cos  6,  about 
its  base. 


142       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

12.  Find  the  volume  of  the  solid  generated  by  the  revolu- 
tion of  the  cycloid  x  =  r  —  r  •  cos  6,  y  =  r  •  6  +  r  •  sin  0,  about 
its  axis. 

13.  Find  the  length  of  the  parabola  y2  —  2px  from  sc  =  0 

to  x  =  x'. 

14.  Find  the  length  of  the  spiral  r  —  a  •  6  from  r  =  0 
to  r  =  r'. 

15.  Find  the  area  bounded  by  the  hyperbola  x2  —  y2  —  a2,  the 
X-axis,  and  the  ordinate  to  the  point  (x,  y)  of  the  hyperbola. 

16.  Show  that  the  area  bounded  by  the  hyperbola  x2—y2=a2, 
the  X-axis,  and  the  line  from  (0,  0)  to  the  point  (x,  y)  of  the 

x  +  y 


hyperbola  is  —  loge— ?- 


Art.  50.  —  Integration  by  Kationalization 

A  derivative  -f-  containing  the  binomial  a  -f  bx  affected  by 
dx 

fractional  exponents  may  be  transformed  into  a  rational  de- 
rivative -2  by  the  substitution  a  +  bx  =  zn,  where  n  is  the 
dz 

least  common  multiple  of  the  denominators  of  the  fractional 
exponents. 


Example  I.  —  Integrate  -f-  = 


j 


dX  rj>$    gj$' 

dx      a   k   dv     dy  dx        6  26  6  z4 


Substituting  «=*•,  ^=6**,  &=&.«5=-Ji_  = 


dz  dz      dx  dz      z3  —  z2     z  —  1 

=  6z3+6z2+6*+6+ JL- 
z—1 


Integrating,  y  =  f  zA  +  2zs  +  3z2  +  6z  -f  61og(z  -  1)  +  C; 
whence,  y  « |  a£  -f  2  a?*  -f  8  x*  +  6  «*  4-  6  log  («*  -  1)  +  (7. 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS     143 
Example  II. — Integrate  -^  = 

**    (i  +  xy  +  (i  +  xy 

Substituting  l+*  =  *°,  |=2,  g«|.g=£_=    *_. 

Integrating,  y  =  2  tan-1z  -f  (7;  whence,  y  =  2  tan-1  (1  +<c)*4-  (7. 

A  derivative  -^  containing  only  the  surd  Va  +  bx  +  x2  may 
da;  r7 
be  transformed  into  a  rational  derivative  —  by  the  substitu- 
dz    J 

tion  Va  -f  bx  -f-  cc2  =  z  —  x. 


dy 


A  derivative  -^  containing  only  the  surd  Va  +  bx  —  x2  may 

(XX  -j 

be  transformed  into  a  rational  derivative  ^  by  the  substitu- 

dz     J 

tion  Va  +  foe  —  x2  =  V(#  —  ^)  (r2  —  «)  =  («  —  rx)  •  2. 


Example  III.  —  Integrate  ^  =  V^as-j-a^ 

da;  a;2 

Writing 

V2^T^  =  z-*,  *  =  -^-,    ^  =  ^+4_z    ,-^*±2*. 
2z+2'    dz       2z  +  2'  2z  +  2 

Hence, 

(fy  =  <fy   dx  =  z2+±z  +  4:=       z2  4(g  +  l)_     1        4 

dz     dx   dz       z2(z  +  l)      z2(z  +  l)     z2(z  +  l)     z  +  1     z* 

Integrating, 


?/=log(z+l)--  +  C=log(tf+l  +  V2a+z2) +C. 

*  X+V2X+X2 

Example  IV.  —  Integrate  ^  =  —      1 

dx     x  V2  +  x  -  x2 


Writing    V2  +x  -  x2  =  V(2  -  x)  (1  +  x)  =  (2  -  *)  .  z, 

T_2*2-l    ^  =  _6z_    and  dy=      2 
"  z2  +  *-'    cto      (z2  +  l)2'  dz     2z2-l 


144       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

By  partial  fractions,  -^  =  — = 

dz      2V2-1      2V2  +  I 

Integrating,  y  =  -L  log  ( V2  ■  2  - 1) L  log  ( V2  .  a  + 1)  +  0. 

V2  V2 

PROBLEMS 
Integrate, 

da?  ^»     ajt  +  i 

2     dy  =       a*      m  dy==_J_. 

dx      V«-l  *   da;     ^  4.  a-l 

3.   4=: *-—  <fy      V6^^"2 

*    (1  +  4*)*  8-  £- ~x 


4.    4= I o     dy_ 


9.    ^ 


:« 


dx     aj-vi  +  a?  '   dx     (2 -f- 3  a?  —  2  aj8)* 

<fy  =        1  10    (fy^  1 

da;     aj+VaT^l  '   dx     xVx2+2x-l 


Art.  51.  —  Evaluation  of  Forms  I00,  00 °,  0° 
If  y  =  /(a;)*(x)  =  1°°  when  x  =  a, 

loge  y  =  <f>  (a?)  •  loge/(#)  =  oo-0  when  a;  =  a. 
If  ?/  =/(a;)*(ac)  =  oo°  when  x  =  a, 

loge  2/  —  ♦(*)■  kfc/O5)  =  0  •  00  when  a;  =  a. 
If  ?/  =/(a?)*(x)  =  0°  when  a;  =  a, 

log*  2/  =  <£  («0  *  l°ge/(#)  =  0  •  go  when  a?  =  a. 

The  true  value  of  \ogey  is  found  by  evaluating  the  form 
00  •  0,  and  the  true  value  of  y  becomes  known. 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS     145 

Example.  —  Find  the  true  value  of  (_  +  l)   when  x  =  oo. 
Here, 

y  ==(  -+1 J  =1*  and  logey=x-  loge(-  +1  J  =  oo  •  0  when  x  =  cc. 

a 


Mi  +  0     I  +  1         a 
Hence,    I0&  y  = ^ J-  = r-  = =  a  when  x  =  oo. 

JC  X2        X 

Since  logey  =  a,  2/  =  ea. 

PROBLEMS 

Find  the  true  value  of, 

1.  ( 1  +  ~i  ]  when  x  =  cc.  5.  (1  -f  »#)•  when  a;  =  0. 

_2_  6.   J  cos  (az)  j  cosec2(c*>  when  a; =0. 

2.  xl~x  when  a?  =  1. 

3.  (sin x)™*  when  x  =  0.  7'  Q  when  *  =  °- 

4.  (sinxVanat  when  #  =  -•  n    ,       n  \\     r.  a 
v        y                        2            8.  (cos  2  a,-)*2  when  a?  =  0. 

9.  (log  a;)*-1  when  x  =  1. 


Art.  52.  —  Differentiation  of  Exponential  Functions 

The  logarithm  of  the  exponential  function  au  is 

logeau=  u-\ogea. 
Differentiating, 

I .  Aa«  =  l0gea  .  f* ;  whence,  — aw  =  aw  .  log,  a  •  — • 
aM    da;  dx  dx  dx 

That  is,  the  derivative  of  an  exponential  function  with  a  con- 


146       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 

stant  base  is  the  product  of  the  exponential  function,  the 
logarithm  of  the  base,  and  the  derivative  of  the  exponent. 

If  a  =  e,  — eu  =  eu  •  — 
dx  dx 

Example.  —  Differentiate,  y  =  of. 

Here  -^  =  of  •  log,  a  •  —  x2  —  2  x  •  of  •  loge  a. 


PROBLEMS 
Differentiate, 

1.1 
1.  y  =  a*.  2.  2/  =  ai-».  3.  y  =  asinx.          4.  y  =  at&n  x. 

5.  y  =  eax.  6.   y  =  e(1+x2).  7.  y  =  e9in~lx. 

8.  Show  that  —  eax  =  aM  •  eax. 
dxu 

Differentiate,   9.  y  =  af .     Here  log?/  =  a; •  logo?,  and  by  dif- 
ferentiation 1 .  ty-  -  i  +  log  #.     Hence,  c^  =  tf  ■  (1  +  log  aj). 

10.  2/  =  eeZ.  11.  i/  =  exX.  12.  y  =  xx*. 

Find  the  true  value  of, 


e*_e-*_2a; 
a;  —  sin  x 


13.    (e*  +  l)x  when  a>  =  0.  16-   "       ~r-  ~  when  x  =  0. 


14'   ^"  Wh6n  *  =  °-  17.   X-   when  z  =  ao. 


15. when  x=a.  ,  0     «- 


(a;  —  a)*  1^*   ex '  s^n  *  when  a;  =  0. 


19.    e"      e  *  when  jj  =  o 

sma; 
i 

20.  Show  that  xx  is  a  minimum  when  x  =  e. 

21.  Find  the  least  value  of  aekx  +  &e~*x. 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS      147 

Art.  53.  —  Integration  of  Exponential  Functions 
The  results  of  the  preceding  article  may  be  written 

J        dx     logea  J       dx 

Example.  —  Integrate  -f-  =  e3*-2. 
dx 

Writing  u  =  Sx-2,    —  =  3    and    ^  =  4^  •— (3  a;-  2). 
dx  dx  dx 

Integrating,  y  =  \  •  e3x~2  -}-  O. 

PROBLEMS 

Integrate, 

1.  ft.£  4.   ^  =  x.<?\  7.   *>  =  a\ 
dx                                 dx  dx 

2.  <k  =  e*.  5.   ^  =  ar<.e*3.  8.   ^  =  a~. 
da;                                 da;  da; 

3.  ^  =  e~.  .         6.    *  =  *  -«.^  9.    &  =  <*•«. 
c?a;                                 da;  da; 

10.   ^  =  a3-5*.  11.   ^  =  e*.sina;. 

da;  da; 


Applying  the  formula  for  integration  by  parts  by  writing 

dv 
dx 


'■,  -^-  =  ex,  y  =  I  ex  •  sin  a;  =  ex  •  sin  a;  —  I  ex  •  cos  a;. 


Applying  the  formula  to   I  ex  •  cos  a;    by  writing    u  =  cos  a;, 
-^  =  ex,  J  ex  •  cos  a;  =  —  ex  •  cos  x  +  I  ex  •  sin  x, 

Hence  I  ex  •  sin  x  =  ez  •  sin  a;  +  ex  •  cos  a;  —  I  e*  •  sin  a-, 

and  y  =  I  e1  •  sin  a?  =  ^-  e*  •  sin  a;  +  \ex  •  cos  a;  +  C. 


148       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


12.    ®  = 

dx 


e 

•  COS  X. 

13. 

dy  _ 
dx 

:  e*  •  a;2. 

14. 

dy 
dx 

5. 

dx 

e> 

>x\ 

16. 

dx 

=  e"" 

-Xs. 

Art.  54.  —  The  Hyperbolic  Functions 

The  functions  \  (e*  +  e~x)  and  J  (ex  —  e-z)  are  called  the 
hyperbolic  cosine  of  x  and  the  hyperbolic  sine  of  x  respectively, 
and  are  denoted  by  the  symbols  cosh  x  and  sinh  x.  Hence  by 
definition  cosh  x  =  \  (ex  +  e-1),  sinh  x  —  \{ex  —  e~x).  It  follows 
at  once  that  (cosh  x)2  —  (sinh  x)2  =  1. 

The  inverse  hyperbolic  functions  are  denoted  by  cosh-1  x  and 
sinh-1#,  so  that  y  =  cosh-1  x  and  y  =  smh.~1x  are  equivalent 
to  #  =  cosh?/  and  x  =  sinh?/  respectively. 

The  hyperbolic  functions  have  been  calculated  and  tabulated. 

Example  I.  —  Find  the  derivative  ofy  =  sinh  x. 
Differentiating, 

^= 4-  sinh  x  =  4-i (e*  -  e-*)  =  I  (e*  +  e-*)  =  cosh x. 
dx     dx  dx2K  /      2v  / 

Example  II.  —  Differentiate  y  =  sinh-1  x. 
Writing  x  =  sinh  y  and  differentiating, 


1  =  cosh  y  •  -^,    whence    —  = 


dx  dx     cosh?/      Vl  +  sinh2?/      yi  +  ar' 


Hence  — sinh-1  a; 


*»  VI  +  a* 

PROBLEMS 

1.  Show  that  —  cosh  x  =  sinh  x. 

dx 

2.  Show  that  —sinh-1-  =  ■ 


dx  a      y  a2  _f-  x2 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS     149 


3.    Show  that  A  cosh"1 - 
dx 


1 


a     y^2 


a~ 


4.  Find  the  minimum  value  of  cosh  a;. 

5.  An  inextensible,  perfectly  flexible,  homogeneous  string 
fastened  at  two  points  in  the  same  horizontal  is  acted  on  by 
gravity  only.     It  is  shown  in 

mechanics  that  for  any  point 
(x,  y)  in  the  position  of  equi- 
librium -2-  =  -,  where  s  is  the 
ax     c 

length  of  string  from  the  low- 
est point  to  (x,  y),  and  c  is  the 
length  of  string  whose  weight 
equals  the  tension  at  the  low- 
est point.  Find  equation  of 
position  of  equilibrium. 

The  ^-derivative  of  ^  =  - 


dx 


ds 


is 


S-t  -hM5 whence 


Fig.  48. 

d  dy 
dx  dx 


Integrating,  sinlr1^  =  *  -f  Cx. 


dx 


When  a>  =  0,^  = 
dx 


0,  hence 


dy 


Ci  =  0,  and  -*-  =  sinh  -.    Integrating  again,  y  =  c-  cosh  -  +  C2. 
dx  c  c 


In   the   figure,   for  x  =  0,   y  =  c,   hence   C2 

y 


0,   and    finally 


c  •  cosh-,  the  equation  of  the  catenary.     Using  exponential 
c 


functions, 


2/  =  |(*c  +  0- 


6.  Find  the  length  of  the  catenary  y  —  c-  cosh  -  from  x  —  0 

c 
to  x  —  x'. 

7.  Find  the  radius  of  curvature  of  the  catenary. 


150       D1FFE11ENTIAL  AND  INTEGRAL   CALCULUS 


e~x  •  dx 


Art.  55.  —  The  Definite  Integral    I 

%J—  oo 

J»+  00 
e-*2  •  dx   is   very   important    in 
-  00 

mathematical   physics   and  the   theory   of    probability.      Its 
value  may  be  determined  by  the  following  analysis,  due  to 

Poisson. 

X+00     , 
e~x  •  dx, 

e~yi  •  dy,  where  x  and  y  are  assumed  to  be  indepen- 
dent variables.     Hence, 

e~**  >dx>  I      e-S  -dy=\  e~^ + "2)  ■  dfe  cZy. 

■oo  %J — a>  */y  =  —  oo   %Jx-=  —  00 


Now,  2;  =  e-(**+»")  represents  the  surface  of  revolution  whose 
generatrix   in  the  ZX-plane   is  2;  =  e-*2  and  axis   of    revolu- 

Jr»y  =  +  oo     •»x  =  +  ao 
e-(*2+y*Kdxdy  is   the 
y  =  —  00     c/ x  =  —  00 


tion  the  Z-axis.     Hence 


LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS     151 

volume  bounded  by  this  surface  of  revolution  and  the 
XF-plane. 

An  element  of  the  volume  of  this  solid  is  the  cylindric  shell 
of  thickness  dr,  radius  r,  altitude  e~(x2+y2)  =  e~''2.     Hence  the 

volume  of  the  solid  is  2  it  I  e~r~  •  rdr  =  ir.  Therefore,  A2  =  -n- 
and  A  —  |       e~x*  •  dx  —  ^/tt. 

To  determine  (    "  e-(*2+2a*) .  ^  wrjte  ^2^2  «#=  (a+a)2-a2; 

«y  —  00 

whence,   (  +  °V(*2  +  2«x) .  dx  _  ^  f  "^-(a+a)2^.^  =  ^-  ,  ^ 

»/  —  00  c/ — 00 

Art.  56.  —  Differentiation  of  a  Definite  Integral 

/*3 
Denoting  the  definite  integral    I    a2  •  x2  •  cfcy  by  ^1,  ^4  =  3  a2 

cZ^l  * 

and  - —  =6  a.     Differentiating  a2  •  x2  •  c?a?  with  respect  to  a, 
da 

and  integrating  the  result  with  respect  to  x  between  the  limits 

—  (a2x2dx)  =  I    2  ax?dx  =  6  «.    Hence  for  this 

d    C3  r3  d 

special    definite    integral,    —  I    a2  •  x2  •  dx  =   I    — (aWdx). 

da  Jo  Jo  da 

That  is,  to  differentiate  the  definite  integral  with  respect  to  a 
parameter,  differentiate  the  function  under  the  sign  of  integra- 
tion with  respect  to  the  parameter. 

To  prove  this  proposition  in  general,  consider  the  definite 
integral  A  =  I  f(x,  a)  dx.  If  f(x,  a)  is  a  continuous  function 
of  a,    I    f(x,  a)  dx  is  also  a  continuous  function  of  a.     For, 

•  / c 

denoting  by  A^L  and  Aa  the  corresponding  changes  of  A  and  a, 
\f(x,  a  -f-  Aa)  —f(x,  a)  \  dx  =  e  I  da  =  e  (6  —  c),  where 
c  is  a  quantity  which  approaches  zero  when  Aa  approaches 
zero.  Hence  A  is  a  continuous  function  of  a.  The  deriva- 
tive of  A  with  respect  to  a  is 


152       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 
d_A  _  Hmit  p/fo«+Ag)--/(s,a)      =  /-  d_  , 

which  proves  the  general  proposition. 

By  this  proposition  the  values  of  definite  integrals  of  a 
general  form  may  be  found  from  known  special  definite 
integrals. 

J"  00  "I 

e~ax  *dx=  — 
o  a 

Differentiating  with  respect  to  a, 

a?  -  e~ax  •  dx  =  ±— -, 
a3 

'xn.e-ax.dx=  1'2'3>'~nm 


Example  II.  —  The  definite  integral   f "  _^_  =  2T.  i, 

5      Jo    a^  +  a2     2  a 
Differentiating  with  respect  to  a} 

J™       dx       ^tt  1  1  r™       dx       =7T  113 

o    (a;2  +  a2)2     2^*2'  Jo    (a2  +  a2)3     2Y2*4'*"' 

da;  tt113     2n-l 


•»oo 

j     as  •  e-°* 

•  da;  = 

1 

V 

X 

j     ic3  •  e-fla 

:.a*a;  = 

1.2-3 

~      a4     '" 

-jf 

X' 


(a^  +  a2)n+1     2  a  2  4         2n 


Art.  57.  —  Mean  Value 


Let  it  be  required  to  determine  the  mean  or  average  value 
of  the  continuous  function  f{x)  from  x  =  a  to  x  =  b.  This  is 
equivalent  to  finding  the  mean  ordinate  of  the  curve  y  =/(#) 
from  x  —  a  to  x  =  b.  Divide  the  portion  of  the  X-axis  from 
x  =  a  to  x  =  b  into  n  equal  parts  Aa;,  and  draw  an  ordinate  of 
yz=zf(x)  at  the  end  of  each  Aa;  nearest  the  origin.  Denot- 
ing the  sum  of  these  ft  ordinates  by  2y,  their  mean  value  is 

— ^  =  — &-! This  is  true  for  all  values  of  n.     When  n  is 

n        n  •  Aa; 


LOGARITHMIC  AND  EXPONENTIAL   FUNCTIONS     153 

indefinitely  increased  Ax  becomes  dx,  %  becomes  the  sum  of 
all  the  ordinates  of  y  =/(#)  from  x  =  a  to  x  =  b,  and  the  mean 

X6                         S*b 
ydx       I    ydx 
—  =  *y >  since 

n  •  Ax  =  6  —  a  for  all  values  of  n. 

Example.  —  Find  the   mean   ordinate   of   the   sine  curve 
y  =  sin  x  from  x  =  0  to  x  =  tt. 

J     sin  ic  •  dx     q 
Here  mean  ordinate  =  *& =  — 

PROBLEMS 

1.  Find  mean  ordinate  of  circle  x2  +  y2  =  r2  from  x  =  +  7  to 
#  =  —  r. 

2.  Find  mean  ordinate  of  ellipse  — +  ^-  =  1  from  #  =  +  a 

a-     b2 
to  x  —  —  a. 

3.  Find  mean  value  of  sin2#  from  x  —  0  to  x  =  7r. 


CHAPTER   IX 

CENTER  OP  MASS  AND  MOMENT   OP  INERTIA 

Art.  58.  —  Center  of  Mass 

If  the  mass  of  a  body  is  divided  into  infinitesimal  elements 
of  mass  dm  and  the  coordinates  of  dm  are  x,  y,  z,  the  center 
of  mass  of  the  body  is  the  point  (x,  y,  z)  so  situated  that  x 
multiplied  by  the  entire  mass  of  the  body  is  equal  to  the  sum 
of  the  products  of  each  element  of  mass  dm  by  the  distance 
of  this  element  of  mass  from  the  YZ-plane,  with  like  defini- 
tions for  y  and  z.      Hence  the  coordinates  of  the  center  of 

I  xdm  I  y dm  I  zdm 

mass  of  a  body  are  x  =  — ,   y  =  — ,    z  =  *— ,  the 

I  dm  I  dm  I  dm 

integration  extending  over  the  entire  mass  of  the  body.* 

Eepresenting  the  magnitude  of  the  element  of  mass  dm  by 
dMy  its  density  by  D,  dm  —  D  •  dM,  and  the  coordinates  of  the 
center  of  mass  become 

*  In  this  chapter  differentials  are  used  directly.  Lagrange  says  in  the 
preface  to  his  Mecanique  Analytique  (1811),  "When  we  have  properly 
conceived  the  spirit  of  the  infinitesimal  method,  and  are  convinced  of  the 
exactness  of  its  results  by  the  geometrical  method  of  prime  and  ultimate 
ratios,  or  by  the  analytical  method  of  derived  functions,  we  may  em- 
ploy infinitely  small  quantities  as  a  sure  and  valuable  means  of  abridging 
and  .simplifying  our  demonstrations." 

154 


CENTER    OF  MASS  AND  MOMENT  OF  INERTIA      155 


Cx  DdM  fy  DdM  Cz  DdM 

CDdM  C  DdM  (%DdM 


the  integration  extending  over  the  entire  magnitude  of  the 
body. 

If  the  body  is  homogeneous,  D  is  constant,  and 

CxdM  fydM  CzdM 

x=^- ,    y  —  ^- ,    z  =  ^- 

CdM  CdM  CdM 

The  center  of  mass  now  becomes  the  center  of  figure.  If 
the  YZ-plane  is  a  plane  of  symmetry  of  the  body,  to  every 
term  +  x  dM  of  j  x  dM  there  corresponds  a  term  —  x  dM. 
Hence  x  =  0 ;  that  is,  the  center  of  mass  lies  in  the  plane 
of  symmetry  of  the  body.  In  like  manner  it  is  shown  that 
the  center  of  mass  lies  in  the  axis  of  symmetry  of  the  body 
and  at  the  center  of  symmetry  of  the  body.  Unless  otherwise 
specified,  the  density  is  assumed  uniform ;  that  is,  the  body  is 
homogeneous. 

In  mechanics  it  is  proved  the  center  of  gravity  of  a  body 
coincides  with  its  center  of  mass. 


Art.  59.  —  Center  of  Mass  of  Lines 

Example  I.  — Find  the  center  of  mass  of  a  straight  line  of 
length  I,  whose  density  varies  as  the  distance  from  one  end. 

Denoting  by  dx  an  element  of 
the  length  of  the  line,  by  x  the  dis- 
tance of  dx  from  the  end  A,  by  k 
the  density  of  the  line  at  unit's  dis- 
tance from  A,  whence 


&. 


AX 

—t — 

Fig.  50. 


156       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


D      x     _ 


£ 


kx  •  x  dx 


£ 


kx'dx 


_  2  7 


Example  II.  —  Find  the  center  of  mass  of  the  arc  of  a 
Y  circle. 

Take   as   X-axis   the   axis   of 
AS  symmetry  of  the  arc.     The  ele- 

ment of  magnitude  is 

From  the  equation  of  the  cir- 
cle x2  +  y2  =  B2, 


Fig.  51. 


R 


dx __  _y 
dy         x 


Hence  ds  =  —dy,  and  x  = 
x 


«c 


+ J  chord 


dy 


radius  x  chord 


arc 


arc 


PROBLEMS 

1.  Find  the  center  of  mass  of  the  straight  line  of  length  I, 
whose  density  varies  as  the  square  of  the  distance  from 
one  end. 

2 .  Find  the  center  of  mass  of  the  entire  cycloidal  arc. 

3.  Find  the  center  of  mass  of  the  length  of  one  loop  of  the 
lemniscate  r1  =  a2  cos  (2  0),  calling  the  length  of  the  loop  2 1. 

Here  dM  =  ds  =  (^  +  ^*V  •  dO  and  x  =  r  .  cos  0. 
V        M  J 

4.  Find  the  center  of  mass  of  the  quarter  of  the  curve 
x3  +  2/3  =  a»  included  by  the  coordinate  axes. 


CENTER   OF  MASS  AND  MOMENT  OF  INERTIA     157 


5.   Find  the   center  of  mass   of    the  helix  x  —  a  •  sin  <f>, 
y  =  a  •  cos  <£,   z  =  c  •  <f>,  from  <j>  =  0  to  <j>  =  <£'. 


Here  cOf  =  *  =  /^  +  <^  +  **  V  .  d* 


d$a  '  fZ<^2     d^V 

6.   Find  the  center  of  mass  of  a  triangle. 

Break  up  the  triangle  into  infinitesimal  strips  by  lines 
parallel  to  the  base  at  intervals  clx  measured  on  the  median 
to  the  base,  and  call  the  distance 
from  the  vertex  to  any  strip  meas- 
ured on  the  median  x.  The  median 
is  an  axis  of  symmetry  of  the  triangle, 
and  the  center  of  mass  must  lie  in 
the  median.  Concentrate  the  magni- 
tude of  each  strip  on  the  median,  and 
the  median  becomes  a  line  whose 
density  varies  as  the  distance  from 
the  vertex.     Hence,  k  representing  the  density  at  unit's  dis- 

rkx  •  xdx 


Fig.  52. 


tance  from  the  vertex,  x  = 


£ 


kx  •  dx 


—  21 


Art.  60.  —  Center  of  Mass  of  Surfaces 


Example  I.  —  Find  the 
center  of  mass  of  the  sur- 
face bounded  by  the 
parabola  y2  =  2 px,  the 
F-axis,  and  the  abscissa 
to  the  point  (x0,  y0)  of  the 
parabola. 

The  surface  is  broken 
up  into  strips  of  breadth 
dy  by  lines  parallel  to  the 


(Xo.y0)^ 

Ay 

AM 

r— 

!  „ 

/ 

' 

X 

A 

Fio.  53. 

158       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


X-axis;  each  strip  is  broken  up  into  elements  of  area  dA 
by  lines  parallel  to  the  F-axis  at  intervals  dx.  Hence 
d3I=dA  =  dxdy  and 

P=y°  fX=jrpxdxdy 

_«/y=0      J  x  =  Q 


x  = 


r3u^( 


=  -At  X, 


—  __Jy  =  Q       Jx  = 

y      ,. s*~ 


2p  y  dx  dy 


y=0       Jx-. 


J/1 

ip  dx  dy 


=  !?/o, 


A  a? 


since  the  object  of  the  cc-integration  is  to  sum  up  the  products 
ydxdy  for  the  elements  of  area  dxdy  forming  the  strip,  and 
the  object  of  the  ^-integration  is  to  sum  up  the  strips  forming 
the  given  surface. 

Example  II.  —  Find  the  center  of  mass  of  the  surface  of 
the  cycloid 
x  =  r  —  r  -  cos  6,  y  =  r-0  +  r  •  sin 0. 

The   center  of    mass   lies   on  the 
X-axis  and 

2xydx 
f*  f*(l  -cos  6)  (0  +  sin  6)  sin  0  ■  dd 

Fig.  54.  =  6  r* 

Example  III.  —  Find  the  center  of  mass  of  the  circular 
sector  whose  angle  is  2  0o  and  whose  density  varies  as  the 
square  of  the  distance  from  the  center. 


-Xa; 


CENTER   OF  MASS  AND  MOMENT  OF  INERTIA     159 


Here  it  is  advisable  to  use  polar  coordinates.  Drawing  radii 
at  angular  intervals  dO,  the  sector  is  broken  up  into  infinitesi- 
mal sectors.  Call  distances 
measured  out  from  the  center 
p  and  draw  circles  concentric 
at  A  at  intervals  dp.  Each 
infinitesimal  sector  is  divided 
into  elements  of  area 

dA  =  pdQ-dp 
and  x  =  p-  cos  0. 

Denoting  the  density  at  unit's 
distance  from  center  by  k, 

^  =  £-2andZ>: 
k      1 

C+6°  CRk.p2.p.d6dp  5       6 

J-6o  Jo 

Example  IV.  —  Find  the  center  of  mass  of  the  eighth  of 
Z 


Fig.  55. 


Hence 


Fig.  56. 


the  surface  of  the  sphere  a?  +  y2  +  z2 
coordinate  planes. 


R2  bounded  by  the 


2   r1 


160       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 
Here  dA  =  (l  +g  +  g)***  -  f  <fedy, 

_     Jo  Jo  z     Xy      2    f/H*-&        xdxdy 

hence    a?  = : — ™ —-^s  I     I  = 

■J-t-B  *RJ    J*  ^/R?-x2-y2 

x  dxl  sin-1 sl J 

\         (#2  -  a?8)  Vo 

=  2tt  rxdx=ziK 

2ttRJo  2 

By  the  nature  of  the  problem  x  —  y  =  z. 

PROBLEMS 

1.  Find  the  center  of  mass  of  the  circular  sector. 

2.  Find  the  center  of  mass  of  the  quarter  of  the  circle 
x2  +  y2  =  B2  included  by  the  coordinate  axes. 

3.  Find  the  center  of  mass  of  the  quarter  of  the  ellipse 

x       u 

—  -f  *-  =  1  included  by  the  coordinate  axes. 

a?     b2  J 

4.  Find  the  center  of  mass  of  the  surface  bounded  by  the 
parabola  y2  =  2px  and  the  double  ordinate  to  the  point  (x,  y). 

5.  Find  the  center  of  mass  of  the  circular  segment  bounded 
by  y2  =  2  Rx  —  x2  and  the  double  ordinate  through  (x,  y). 

6.  Find  the  center  of  mass  of  the  surface  bounded  by  the 
circle  y2  —  2  Rx  —  x2,  the  y-axis,  and  the  abscissa  to  the  point 
(x,  y).     This  surface  is  called  the  circular  spandrel. 

7.  Find  the  center  of  mass  of  the  part  of  a  circular  annulus 
bounded  by  the  circles  r  =  R,  r=  R',  and  the  radii  vectores 
0  =  —  00,  0  =  -f-  00. 

8.  Find  the  center  of  mass  of  the  surface  bounded  by 
x*  +  y*  =  a*  and  the  positive  coordinate  axes. 

9.  Find  the  center  of  mass  of  the  area  of  one  loop  of  the 
lemniscate. 

10.    Find  the  center  of  mass  of  a  zone. 


CENTER   OF  MASS  AND  MOMENT  OF  INERTIA      161 


Art.  61.  —  Center  of  Mass  of  Solids 
Example  I.  —  Find  the  center  of  mass  of  the  half  of  the 


X2 

ellipsoid  —  + 


+ 


1  lying 


a"     b2     c 
to  the  right  of  the  Z  F-plane. 
The  X-axis  is  an  axis  of 
symmetry,  and 

dM=  ttvs  -rt-dx 


^{a2-x2)dx. 
a2 


Fig.  57. 


Hence 


-9  C(a?-x2)dx 
a2Jo 


x  = 


■abc 


=  *a. 


Example  II.  —  Find  the  center  of  mass  of  the  eighth  of 
the  sphere  x2  -f  y2  +  z2  =  E2  included  by  the  coordinate  planes, 
the  density  varying  as  the  square  of  the  distance  from  the 
center. 

Here  it  is  advisable  to  use  polar  coordinates.  Passing 
planes  through  the  Z-axis  at  angular  intervals  d<f>,  the  solid  is 
divided  into  spherical  wedges.  Passing  conical  surfaces  with 
vertex  at  0,  and  whose  elements  Z 

make  angles  with  the  X  F-plane 
increasing  by  d6,  each  wedge  is 
divided  into  pyramids.  Passing 
spherical  surfaces  concentric  at 
0,  and  whose  radii  increase  by 
dp,  each  pyramid  is  divided 
into  elementary  parallel  opi- 
peds,  whose  dimensions  are  dp, 
p  cos  6  dd>.  p  dO.     Hence,  Y' 

r  ^;   r  7  Fig.  58. 

dM  =  p2  cos  0  dp  d<f>  dO,   x  =  p  cos  0  cos  <p,   D  —  kp2, 


162       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

where  k  is  the  density  at  unit's  distance  from  the  center. 
Finally, 

IJl     k  •  p5  >  cos20  •  cos  <£  •  d<f>-  d6  -dp 


x  = 


n»2    f*R 
&  y  •  cos  0  •  d<£  •  cZ0  •  dp 


=  **■ 


PROBLEMS 

1.  Find  the  center  of  mass  of  a  hemisphere. 

2.  Find  the  center  of  mass  of  the  paraboloid  of  revolution 
generated  by  revolving  y2  =  2  px  about  the  X-axis,  and  in- 
cluded by  the  planes  x  =  0,  x  =  x'. 

3.  Find  the  center  of  mass  of  the  solid  generated  by  re- 

ar2    v2 
volving  half  the  ellipse  —  +  —2  =  1  from  x  =  0  to  x  =  a  about 

the  X-axis.  a      b 

4.  Find  the  center  of  mass  of  the  rectangular  wedge. 


Art. 


Theorems  of  Pappus* 


:jl 


ds 


Multiplying  both  sides  of  the  equation  y  =  ** by  2  its, 

2  iry  •  s  =  2  7T I  y  •  ds.     The  right-hand  member  of  this  equa- 

Y  3$C 7Y~  ti°n  rePresents  the  area  of 

the  surface  generated  by  the 
revolution  of  the  line  s  about 
the  X-axis;  the  left-hand 
member  is  the  length  of  the 
line  s  multiplied  by  the  cir- 
cumference of  the  circle  de- 
scribed by  the  center  of  mass 
Fig.  50.  of  the  revolving  line. 

*  First  published  by  Pappus  of  Alexandria  about  the  end  of  the  third 
century  of  our  era. 


CENTER   OF  MASS  AND  MOMENT  OF  INERTIA      163 


This  furnishes  a  convenient  determination  of  the  center  of 
mass  of  the  line  when  the  area  generated  is  known,  or  of  the 
area  generated  when  the  center  of  mass  of  the  line  is  known. 

Example  I.  —  Determine  the 
center  of  mass  of  the  circular 
arc. 

The  arc  revolving  about  the 
X-axis  generates  a  zone  whose 
area  is  2  ttR  •  chord.  Hence, 
2  iry  •  arc  =  2ttR  •  chord,     and 

R  •  chord 


Fig.  60. 


y  = 

as  before  found. 


arc 


Example  II.  —  Find  the  area  of  the  surface  generated  by 
revolving  the  cycloid  y  =  r  vers-1-  -f  V2  rx  —  x2  about  its  base. 

The  distance  of  the  center  of  mass  of  the  cycloidal  arc  from 
the  base  is  f  r,  the  length  of  the  cycloid  is  8  r.  Hence, 
area  =  2  tt  •  %  r  •  8  r  =  -\4-  -n-r2. 


PROBLEMS 

1.  Find  the  area  of  the  surface  generated  by  the  revolution 
about  the  X-axis  of  the  circle  of  radius  r,  distance  of  center 
from  X-axis  a,  where  a  >  r. 

2.  Find  the  area  of  the  surface  generated  by  the  revolution 
of  a  semi-circumference  of  radius  r  about  a  tangent  at  its 
middle  point. 

Multiplying  both  sides  of  the  equation, 
Cyydxdy     \\yLdx 

A 


y  = 


■    I  dxdy 


by  2  ir  A,  2  iry  -A 


—J> 


dx. 


164       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 


The  right-hand  member  of  this  equation  represents  the 
volume  of  the  solid  generated  by  the  revolution  of  the  area 
A  about  the  X-axis;  the  left-hand  member  is  the  area  A 
multiplied  by  the  circumference  described  by  the  center  of 
mass  of  A.  This  furnishes  a  convenient  determination  of  the 
center  of  mass  of  the  area  when  the  volume  generated  is 
known,  or  of  the  volume  when  the  center  of  mass  of  the  area 
is  known. 


Fig.  61. 

Example  I.  —  Find  the  volume  generated  by  the  revolution 
about  the  X-axis  of  the  ellipse  whose  axes  are  2  a  and  2  b, 
distance  of  center  from  X-axis  c.     V  =  2  ire  •  -n-ab  =  2  nhtbc. 

Example  II.  —  Find  the  center 
of  mass  of  a  circular  sector. 

The  volume  generated  by  the 
circular  sector  whose  angle  is 
2  Oq,  radius  R,  revolving  about  a 
diameter  parallel  to  the  chord  of 
the  sector  is  2  -n-R  •  2  R  sin  0O  •  \  R, 
the  area  of  the  sector  is  R2  •  0o. 

Hence,       y-|SgA: 

Fig.  62.  3        0Q 


CENTER   OF  MASS  AND  MOMENT  OF  INERTIA      165 


PROBLEMS 

1.  Find  the  volume  generated  by  revolving  the  cycloid 
y  =  r  •  vers-1-  -f-  V2  rx  —  ar  about  its  base. 

2.  Find   the   distance  of  the  center  of  mass  of  half   the 

x2     v2 
ellipse  '--  +  •—„  =  1  bounded  by  x  =  0,  x  =  a,  from  the  center 
a2      b2 

of  the  ellipse. 

Art.  63.  —  Moment  of  Inertia 

If  the  mass  of  a  body  is  divided  into  infinitesimal  elements 
of  mass,  and  each  element  is  multiplied  by  the  square  of  its 
distance  from  a  fixed  line,  the  sum  of  all  these  products  is 
called  the  moment  of  inertia  of  the  body  with  respect  to  the 
straight  line  as  axis.  Denoting  the  infinitesimal  element  of 
mass  by  dm,  its  distance  from  the  axis  by  y,  and  the  moment 
of  inertia  by  I,  /  =  I  y2  -  dm,  the  integration  extending  over 
the  entire  mass  of  the  body. 

Representing  an  infinitesimal  element  of  the  magnitude  of 
the  body  by  dM,  the  density  of  this  element  by  D,  dm—D  -  dM, 
and  the  moment  of  inertia  becomes  I  =  J  y2  •  D  •  dM,  the  in- 
tegration extending  over  the  entire  magnitude  of  the  body. 

The  moment  of  inertia  occurs  very  frequently  in  Strength  of 
Materials  and  in  the  Theory  of  Eotation  of  Bodies.  For 
example,  in  Problems  31,  32,  33  of  Article  23,  i"  is  the  moment 
of  inertia  of  the  cross-section  of  the  beam. 

If  the  material  of  the  body  is  of  uniform  unit  density,  which 
is  always  assumed  unless  the  contrary  is  stated,  D-dM=  dM, 
and  the  moment  of  inertia  depends  only  on  the  shape  of  the 
body. 

The  moment  of  inertia  is  one  of  the  elements  used  in  select- 
ing shapes  for  engineering  structures. 


166        DIFFERENTIAL   AND  INTEGRAL    CALCULUS 


The  moment  of  inertia  divided  by  the  mass  of  the  body  is 
called  the  square  of  the  radius  of  gyration  of  the  body  for  the 
moment   axis  used   and   is    denoted 
I  y'1  dm 


by  k2.     Hence  k2  =' 


k  is  the 


distance  from  the  moment  axis  to  the 
point  where  the  mass  of  the  body 
must  be  concentrated  so  that  the 
moment  of  inertia  of  the  concen- 
trated mass  shall  equal  the  moment 
of  inertia  of  the  mass  distributed 
throughout  the  body.  In  Problem 
35  of  Article  23  and  in  Problem  13  of  Article  43,  k2  is  the 
square  of  the  radius  of  gyration  of  the  turning  body. 

Denote  by  Ia  the  moment  of   inertia  of   a  body  for   any 
moment  axis  AA,  by  Ic  the  moment  of  inertia  of  the  same 

body  for  a  parallel  moment  axis 
through  the  center  of  mass  of 
the  body.  Through  C,  the 
center  of  mass  of  the  body,  pass 
a  plane  US  perpendicular  to 
the  line  AC.  Denote  by  D  the 
distance  between  the  two  axes, 
by  r  the  distance  of  dM  from 
the  axis  AA,  by  r'  the  distance 
of  dM  from  the  axis  CO.     By 

:  I  r'2  •  dM.     From    the    triangle 


Fig.  64. 


-S 


definition  Ia  =  f  r2  •  dM,  Ic 


ACP,  r2  =  r' 2  +  D2  +  2  r'  D  sin  6.     Hence 

(1)        Cr2  dM  =  Cr'  2dM+D2  CdM  +  2  D  Cr '  sin  6 .  dM. 


CENTER   OF  MASS  AND  MOMENT  OF  INERTIA      167 

Now        r'  sin 6  =  PE  =  y  and    I  r' sin6-dM=  i y  dM. 

CydM 
But  y  = ^ —  =  0,  since  y  is  the  distance  of  the  center  of 

mass  of  the  body  from  the  plane  RS,  and  this  plane  is  drawn 
through  C  perpendicular  to  the  line  CA.  Hence  (1)  becomes 
Ia  =  Ic  +  D2-  M,  which  may  be  written  Ic  =  Ia  —  D2-  M,  and  is 
known  as  the  reduction  formula.  Stated  in  words,  this  formula 
reads  the  moment  of  inertia  of  a  body  for  any  moment  axis 
equals  the  moment  of  inertia  for  a  parallel  moment  axis  through 
the  center  of  mass  of  the  body  plus  the  square  of  the  distance 
between  the  two  axes  into  the  mass  of  the  body. 

Denoting  the  radii  of  gyration  for  the  axis  AA  and  the 
axis  CC  by  Ka  and  Ke  respectively,  Ia  —  M-  Ka2,  Ic  =  M-  K2, 
and  by  the  reduction  formula  K2  =  K2  +  D2. 

Art.  64.  —  Moment  of  Inertia  of  Lines  and  Surfaces 

Example  I.  —  Find  the  moment  of  inertia  of  a  straight  line 
of  length  I  for  moment  axis  perpendicular  to  line  through  one 
end  of  line. 

Denoting  an  element  of  the  line 
by  dy,  the  distance  of  this  element 
from  the  moment  axis  by  y, 

The  radius  of  gyration  is  found  from 

A 

7  o        -o  I         .  79  Fig.  65. 

k2=^-=\l\ 

The  moment  of  inertia  for  a  parallel  axis  through  the  center 
of  mass  of  the  line  is  Ic  =  ^  Z3  —  \  Is  =  TJY  Z3.  The  radius  of 
gyration  for  this  axis  is  found  from  ft2  =  -^ I2. 


-V- 


AV 


168        DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Fig.  67. 


Example  II.  —  Find  the  moment  of  inertia  of  a  triangle  for 
moment  axis  through  vertex  parallel  to  base  of  triangle. 

Breaking  up  the  triangle  into  strips  by  lines  parallel  to  the 
moment  axis  at  intervals  cly,  calling  the  length  of  any  strip  x, 

and  its  distance  from  the  moment 
axis  y,  dM  =  x  dp,  and  from  simi- 
lar triangles  -  =  —    Hence, 

The  radius  of  gyration  is  found 

from  *=!£=**• 

The  moment  of  inertia  for  a  parallel  axis  through  the  center 
of  mass  of  triangle  is 

Ic  =  J  M3  -  ±bd  (j  d)2  =  ^  bd3  and  kc2  =  TV  d\ 

The  moment  of  inertia  for  the  base  of  triangle  as  axis  is 
Ia  =  h  bd*  +  it>d($  df  =  tl  bd*  and  kj  =  |  d\ 

Example  III.  —  Find  the  moment  of  inertia  of  a  circle  for 
axis  through  center  perpendicular  to  plane  of  circle. 

Y  Here    it   is    advisable   to  use 

polar  coordinates.  dM  =  pdOdp, 
and  the  distance  of  dM  from 
the  axis  AP  is  p.     Hence, 


J     p3d0dp  =  ±7rlt*, 


and  kp2  =  ±R2. 


Fig.  68. 


Ip  is  called  the 
polar  moment  of  inertia  of  the 
circle. 

Denoting  by  Ix  the  moment  of 


CENTER   OF  MASS  AND  MOMENT  OF  INERTIA      169 

inertia  of  the  circle  for  axis  AX,  by  Iy  the  moment  of  inertia 
for  axis  AY, 

Ix  =  fy*dM,   Iy=Cx*dM, 

and  Ix  +  Iy  = jV  +  f)  dM  =§P*dM  =  Ip. 

Since  the  circle  is  placed  in  exactly  the  same  manner  with 
respect  to  each  of  the  diameters  AX  and  AY,  Ix  =  Iy.     Hence, 

2/a;  =  ip  =  |7ri24,  and  Ix  =  {ttR\  k*=\R\ 

PROBLEMS 

1.  Find  moment  of  inertia  of  a  rectangle  base  b,  altitude  d 
for  base  of  rectangle  as  axis. 

2.  From  Problem  1  find  by  reduction  formula  the  moment  of 
inertia  of  rectangle  for  axis  through  center  of  rectangle  parallel 
to  base. 

3.  Find  moment  of  inertia  of  rectangle  for  axis  through 
center  perpendicular  to  plane  of  rectangle. 

4.  Find  moment  of  inertia  of  isosceles  triangle  for  axis  of 
symmetry  as  moment  axis. 

5.  Find  moment  of  inertia  of  ellipse  for  major  diameter  as 
moment  axis. 

6.  Find  moment  of  inertia  of  ellipse  for  minor  diameter  as 
moment  axis. 

7.  Find  moment  of  inertia  of  ellipse  for  axis  through  center 
of  ellipse  perpendicular  to  plane  of  ellipse. 

8.  Find  moment  of  inertia  of  parabolic  segment  for  axis  of 
parabola  as  moment  axis. 

9.  Find  moment  of  inertia  of  circular  spandril  for  diameter 
as  moment  axis. 


170       DIFFERENTIAL  AND  INTEGRAL    CALCULUS 


Art  65.  —  Moment  of  Inertia  of  Solids 

Example  I.  —  Find  the  moment  of  inertia  of  the  rectangular 
parallelopiped  whose  dimensions  are  b,  d,  h  for  axis  through 
centers  of  two  opposite  faces. 

Break  up  the  solid  into  laminae  by  planes  perpendicular  to 
the  axis  at  intervals  dx  and  call  the  distance  of  any  lamina 


Sftt 

H-1- 


-b*4 


Fig.  69. 


from  one  of  the  faces  x.  The  lamina  may  be  broken  up  into 
elements  dxdydz,  and  the  moment  of  inertia  of   the  lamina 

=  dx  I  I  p2dydz  =  moment  of  inertia  of  base  of  lamina  mul- 
tiplied by  the  thickness  of  the  lamina.  Hence  the  moment 
of  inertia  of  the  lamina  is  ^(b2+d2)bddx,  and  the  moment  of 
inertia  of  the  parallelopiped  is 

I ■=  TL  (p2  +  d2)  bdC  dx  =  ^{b2  +  d2)  bdh. 

The  radius  of  gyration  is  found  from 


&°=^ 


(b1  +  d2)  bdh 


bdh 


=  TV(&8  +  <P). 


Example  II.  —  Find  the  moment  of  inertia  of  a  cone  of 
revolution  for  axis  through  center  of  mass  of  cone  parallel  to 
base  of  cone. 


CENTER   OF  MASS  AND  MOMENT  OF  INERTIA      171 

Break  up  the  cone  into  laminae  by  planes  parallel  to  the 
base  at  intervals  dy.  Call  the  distance  of  any  lamina  from  the 
vertex  y,  the  radius 
of    the   base   of    the  T 

lamina  x.  The  mo- 
ment of  inertia  of  the 
lamina  for  axis  X'  X' 
through  the  center  of 
the  lamina  and  paral- 
lel to  the  axis  XX  is 
\irx*dy,  the  distance 
from  X'X'  to  XX  is 
\H—y.  Hence  by  the  reduction  formula  the  moment  of 
inertia  of  the  lamina  for  axis  XX  is  \irx*dy  +  irxP^H—yydy, 
and  the  moment  of  inertia  of  the  entire  cone  is 

I=f*\\«*dy  +  Tr^Cftf-  yfdyl . 
By  similar  triangles    ~  = — ,  hence 


Fig.  70. 


PROBLEMS 

1.  Find  moment  of  inertia  of  the  sphere  for  diameter  as 
axis. 

2.  Find  moment  of  inertia  of  cone  of  revolution  for  axis 
of  symmetry  as  moment  axis. 

3.  Find  moment  of  inertia  of  cylinder  of  revolution  for 
axis  of  symmetry  as  moment  axis. 

4.  Find  moment  of  inertia  of  cylinder  of    revolution  for 
axis  through  center  parallel  to  base. 


172       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

5.  Find  moment  of   inertia  of   spherical  cap  for  axis  of 
symmetry  as  moment  axis. 

6.  Find  moment  of  inertia  of  ellipsoid  for  longest  diameter 
as  moment  axis. 

7.  Find  moment  of  inertia  of  the  segment  of  a  paraboloid 
of  revolution  for  axis  of  symmetry  as  moment  axis. 


CHAPTER   X 

EXPANSIONS 

Art.  66.  —  Convergent  Power  Series 

The  identical  equation  (1  —  x)'2=  1  —  2  x  +  x2  is  true  for  all 
values  of  x. 

Expanding  the  fraction into  a  series  by  division,  there 

jl  —  x 

results  the  identical  equation 

-5L  =  i  +  x+x2+xs+xA+  ...xn-1  +  xn+xn+1+xn+2+xn+!i+  .... 
1— x 

where  the  number  of  terms  in  the  series  is  infinite.     This 

identity  is  not  true  for  all  values  of  x.     For  example,  if  x  =  2, 

the  fraction  equals  —  1,  the  series  equals  infinity.     To 

JL  —  X 

determine  for  what  values  of  x  the  identity  is  true,  denote  the 
sum  of  the  first  n  terms  of  the  series  by  sn,  the  sum  of  the 

remaining  terms  by  rn.     Then  =  sn  +  rnt  and  if  rn  can  be 

1  —  x 

made  less  than  any  assigned  quantity,  however  small,  by  tak- 
ing n  sufficiently  large,   =  limit  sn  when    limit  n  =  oo 

1  —  x 
and  the  series  is  said  to  be  convergent.     Now 

rn  =  xn(l  +8  +  arJ  +  arB  +  a4H ). 

When  x  <  1,  rn  =  -1 — ,  and  limit  rn  =  0  when  limit  n  =  oo 
1  —  x  1 

Hence  when  x  <  1,  the  value  of  the  fraction is  correctly 

1  —  x  v 

represented  by  the  infinite  series 

173 


174        DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

(1)       1  +  x  +  x2  +  x?  +  oj4  H af1-1  +  xn  +  #n+1  +  a;n+2  H , 

and  sn  approaches  the  value  of  the  fraction  more  and 

1  —  x 

more  closely  the  larger  n  is  taken.  For  example,  when  x  =  \, 
the  fraction  equals  2,  while  s4  =  *f  =  1.8889,  s8  =  f|f  =  1.9914. 
This  infinite  series  is  convergent  when  —  1  <  x  <  1  and  con- 
verges towards  2  when  a;  =  -J-.  The  totality  of  values  of  x  for 
which  the  infinite  series  is  convergent  is  called  the  region  of 
convergence  of  the  infinite  series. 

=1i 0 ±i 

Fig.  71. 

The  heavy  portion  of  the  straight  line  shows  the  region  of 
convergence  of  the  infinite  series  (1). 
In  general,  the  infinite  series 

a0  +  a,!  > x  +  a2  •  x2  -\-  a3  > x?  +  a4- x4 -\-  •  •  • 

+  «n-i  •  ^n_1  +  an  •  xn  +  an+1  •  xn+1  +  •  •  -, 

where  the  coefficients  are  finite  and  independent  of  x,  is  called 
a  power  series.  Denoting  the  sum  of  the  first  n  terms  of  the 
power  series  by  sn,  the  sum  of  the  remaining  terms  by  rn, 

a0  -f  a  •  x  +  a.2  •  x2  +  a3  •  Xs  -f-  •  •  • 

+  a„_i  •  a;"-1  +  aw  •  xn  +  an+1  ■  xn+1  -f-  ...  =  sn  +  rn. 

For  each  value  of  x  for  which  limit  rn  =  0  when  limit 
n  =  «x,  the  power  series  is  convergent  and  has  a  determinate 
finite  value,  which  is  the  limit  of  sn  when  limit  n  =  oo.  Hence 
within  its  region  of  convergence  the  power  series  defines  a 
function  of  a?  and  may  be  denoted  by /(a). 

A  convenient  test  for  the  convergence  of  a  power  series  is 
Cauchy's  test,  which  reads : 

A  power  series  is  convergent  if  from  and  after  some  fixed 


EXPANSIONS  1 75 

term  the  ratio  of  each  term  to  the  preceding  term  is  always 
numerically  less  than  some  number  numerically  less  than  unity. 
For  suppose  that  from  and  after  the  m  +  2  term,  m  being 
finite,  of  the  power  series 

a0  +  ax  •  x  +  a,  •  x2  -\ am  •  xm  +  am+1  •  xm+1  +  am+2 .  zTO+2  +  ... 

the  ratio  of  each  term  to  the  preceding  term  is  numerically 
less  than  r,  and  that  —  1  <  r  <  1.  Denote  the  sum  of  the  first 
m  terms  of  the  series  by  sOT,  the  sum  of  the  next  n  terms  by 
sn,  the  sum  of  the  remaining  terms  by  rn,  the  sum  of  the 
entire  series  by  s.     Then  s  =  sm  +  sn  +  rn, 

sn  +  rn  =  am  •  xm  +  am+1  •  xm+1  +  am+2  •  xm+2  +  am+3 .  xm+s  +  ... 
<  a„  •  am(l  +r  4- r8  +  f* H ?*n  +  ?'n+1  4-  ?"n+3  H ) 

by   hypothesis.      Hence    rn  <  aw  •  xm- ,   and   limit   rn=  0 

1  —  r 

when  limit  n  =  oo,  since  am  and  a?  are  by  hypothesis  not 
infinite  and  r  is  assumed  less  than  unity.  Hence  the  sum 
of  the  series  s  =  sm  +  limit  sn  when  limit  n  =  oo ;  the  series  is 
convergent  and  Cauchy's  test  is  proved. 

For  example,  in  the  series  l-f-sc-l-a^  +  a^-f-  ..-a;n-f  xn+l -\ , 

the  ratio  of  each  term  to  the  preceding  term  is  x,  and  by 
Cauchy's  test  the  series  is  convergent  when  —  1  <  x  <  1.* 

Art.  67.  —  Taylor's  and  Maclaurin's  Series 

The  values  of  explicit  algebraic  functions  can  be  directly 
calculated  for  arbitrarily  assigned  values  of  the  independent 
variable.  For  example,  if  y  =  x*  —  1  x  +  7,  any  value  may  be 
assigned  to  x,  and  the  corresponding  value  of  y  becomes 
known. 

*  Euler  seems  to  have  been  the  first  to  call  attention  to  the  fact  that 
infinite  series  can  be  safely  used  only  when  convergent. 


176       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Consider  the  function  f(x)  =  xm,  when  m  is  assumed  to  be 
a  positive  integer.     When  x=b  =  a -\- h,  whence  h  =  b  —  a, 

f(b)  =  (a+h)m  =  am  +  m  •  am~l  -h  +  m  (m  -  1)  am-2£- 

+  m(m-l)(m-2)aw-3^-  +  ... 
ol 

by  the  binomial  formula.     This  result  may  be  written, 

(1)  /(b)  =/(«  +  ft)  =/(a)+/'(a).  h  ^ 

+/"(«)  •§+/"<•)  -|+- 
or  (2)  /(&)  =  /(a)  +/(«)•  (6  -a) 

+/"(«)  •  ^fr12  +/"<«)  •  ^=p^ + -, 

where  f(a),  /'(a),  f"(a),  f'"(a),  -~,  are  the  values  of  the  suc- 
cessive derivatives  of  f(x)  —  xm  when  x=  a. 

The  symbol  n\  stands  for  1  •  2  •  3  •  4  •  5  •  •••  n,  and  is  read 
factorial  n. 

It  is  proposed  to  derive  a  general  series  of  the  same  form  as 
series  (1)  and  (2)  for  the  approximate  calculation  of  any  func- 
tion f(x)  for  arbitrarily  assigned  values  of  the  independent 
variable.  The  problem  may  be  thus  formulated :  Let  f(x) 
and  its  successive  derivatives  f'(x),  /"(#),  f'"(x),  •••,  be  finite 
and  continuous  from  x  =  a  to  x=b,  and  denote  by  f(a),  f'(d), 
f"(a),  /'"(a),  ---,  the  values  of  these  functions  when  x  =  a. 
The  value  of  f(b)  is  to  be  found  in  terms  of  /(a),  f(a),  f"(a), 
/'"(a),  ••-,  and  the  powers  of  b  —  a. 

The  investigation  is  based  on  the  following  proposition, 
known  as  Eolle's  theorem : 

If  the  function  cf>  (x)  and  its  first  derivative  <f>'(x)  are  finite 
and  continuous  from  x  =  a  to  x  =  b  and  <f>  (a)  =  0  and  cf>  (b)  =  0, 
the  first  derivative  <j>'(x)  must  vanish  for  some  value  of  x 
between  a  and  b. 


EXPANSIONS  177 

If  <f>'  (x)  is  identically  zero,  the  truth  of  the  proposition  is 
evident.  If  </>'  (x)  is  not  identically  zero,  $'  (x)  must  change 
sign  between  x  =  a  and  x  =  b,  for  otherwise  cj>  (x)  would  either 
continually  increase  or  continually  decrease  from  a;  =  a  to 
x  =  b,  and  in  neither  case  could  </>  (a)  and  <f>  (b)  both  be  zero. 
Hence  <£'  (x)  must  change  sign  between  x  =  a  and  x  =  b. 
Since  <£'  (a;)  is  assumed  to  be  finite  and  continuous  from  x  =  a 
to  x  =  b,  </>'  (x)  can  change  sign  only  by  passing  through  zero. 
Therefore  </>'(#)  must  vanish  for  some  value  of  x  between 
x  =  a  and  x  =  b.  Denoting  this  value  of  x  by  a  +  6(b  —  a), 
where  6  must  be  less  than  unity,  <f>'\  a  +  6(b  —  a) }  =  0. 

Now  assume  f(b)  =  f(a)  +  kx(p  —  a),  where  kx  is  a  constant  to 
be  determined.  Write  (f)l(x)  =  f(b)  —  f(x)  —  kx(b  —  x)  and  form 
the  first  derivation  <£/(#)=  —  f'(x)+  kx.  Since  by  hypothesis 
<f>x(x)  and  <f>i(x)  are  finite  and  continuous  from  x  =  a  to  g  =  6 
and  <f>x(a)=0  and  <£i(&)  =  0,  <£i'(#)  must  vanish  for  some  value 
of  x  between  a  and  b.     Denoting  this  value  of  x  by 

a  +  0x(b  -  a),  kx  =f\a  +  ^(6  -  a)  j. 

Hence        f(b)=f(a)  +f\a  +  0x(b  -  a)\(b  -  a). 

Next  assume    f(b)  =  f(a)+f'(a)  •  (6  -  a)+k2.(b  ~  a)2- 

Write    ^(«)s/(6)- /(«)-/(»).  (6 -a?)-^.^^  and 

form  the  first  derivative  <£</(#)  =  —  (&  —  #)/"(#)  +  fc«(6  —  a>). 
Since  by  hypothesis  4>2(x)  and  <£2'(#)  are  finite  and  continuous 
from  x  =  a  to  a?  =  &  and  <£2(a)=0  and  <£2(&)=0,  <£2'(;c)  must 
vanish  for  some  value  of  x  between  a  and  b.  Denoting  this 
value   of  x  by  a-\-02(b  —  a),  k2—fu\a-\-B2(b  —  a)\.     Hence 

/(6)=/(a)+/'(«)  •  (6  -  «)+/>  +  02(&  -«)f  -^fp^ 


178        DIFFERENTIAL   AND  INTEGRAL   CALCULUS 

Assume 

f(b)=f(a)+f(a)  •  (5  -  a)+f%a)  :£=&  +  h-(J~f£- 

Write  U*)  =  fQ>)-f(x)-f'(x)  -(b-x) 

J       2!  3! 

and  form  the  first  derivative 

Since  <£3(#)  and  <£3'(#)  are  by  hypothesis  finite  and  continu- 
ous from  #  =  a  to  x  =  b  and  <f)S(a)=0  and  <£3(6)=0,  <£3'(a;) 
must  vanish  for  some  value  of  x  between  a  and  6.  Denoting 
this  value  of  x  by  a  +  03(6  —  a),  k3  =f"  { a  -f  03(6  —a)}.     Hence 

f(b)=f(a)+f(a)  •  (6  -  a)  +  /"(a)  .^I2 


(6  -  a)3 


+/■*!« +  «*■- a)  J 

Kepeating  this  operation  n  times,  there  results 

(1)  /(6)=/(«)+/<(a) ■  Q>-a)+f(a).(t=«t  +  r(a)  .(tz^f 

J    W        4 !  7      w    (« - 1) ! 

+f>\a  +  6(b-a)\-(-h -«■)", 

n ! 

where  0  <  6  <  1. 

The  error  committed  by  placing  f(b)  equal  to  the  sum  of 
the  first  n  terms  of  series  (1)  is  rn  =fn\a  +  6(b  -  ft)  j  (b  ~  a^- 

If  this  error  can  be  made  indefinitely  small  by  taking  n  suffi- 
ciently large,  the  series  is  convergent  and  can  be  used  to  cal- 
culate the  value  of  /(d)  to  any  required  degree  of  accuracy, 
provided  /(a),  /'(a),  /"(ft),  /'"(a),  •••   are  known.     When  n 


EXPANSIONS  179 

becomes  indefinitely  large,  the  series  becomes  an  infinite  series 
and  the  region  of  convergence  is  most  readily  determined  by 
Cauchy's  test. 

In  (1)  place  b  —  a  =  h,  whence  b  =  a  +  h.     There  results 

(2)  f(b)  =/(a  +  A)  =  /(«)  +  /'(<*)  •  h  +r(*)~  +/"'(<*)  ~ 

This  is  Taylor's  series.     In  (2)  place  a  =  0.     There  results 

(3)  f(h)  =/(0)  +/'(0)  •  h  +/»(0)  -|+/'"(0)  -| 

This  is  Maclaurin's  series.     The  only  restrictions  on  a  and  h 
in  these  series   are  that  f{x),  f'(x),  f"(x),  f'"(x),  -••fn~x{x) 

must  be  finite  and  continuous  from  x  =  a  to  x  =  a-\-h  and 

hn 
that /"  (a  +  07*) —  must    become    less    than   any   assignable 

quantity  when  n  is  indefinitely  increased.  Since  the  quantities 
represented  by  a  and  h  are  not  fixed,  they  may  be  denoted  by 
x  and  y  respectively,  when  Taylor's  and  Maclaurin's  series 
become 

f(x  +  y)=f(  x)+f(x).  y+f"  (*).£+/»(*).£+.. . 

f(y)  =/(0)  +/'  (0)  •  y  +/"  (0)  -  £  +/"<  (0)  ■£+  - 


*Taylor  (1685-1731)  published  his  series  in  his  "  Methodus  incremen- 
torum."  Maclaurin  published  his  series  in  his  "Treatise  of  Flexions" 
1742.  The  expansions  effected  by  these  series  had  been  previously 
obtained  by  laborious  processes. 


180       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Taylor's  series  expands  a  function  of  the  sum  of  two  varia- 
bles in  the  ascending  powers  of  one  of  the  variables;  Mac- 
laurin's  formula  expands  a  function  of  one  variable  in  the 
ascending  powers  of  that  variable. 

Example  I.  —  Expand  loge  (1  -f  y). 

This  expansion  is  effected  by  Maclaurin's  series  since 
log,.  (1  +  y)  is  a  function  of  one  variable.  Forming  the  succes- 
sive derivatives, 

2! 


fi3f)  =  log.  (1+30,        f'"(y)  = 


(1  +  y) 


f'(y)  =A->  /"&)=■  ' 


hence     /(0)  =  0,    /'(0)  =  1,    /»(0)=-l,    /'"(0)  =  2!, 
jT(0)=-3!,  ...  /»(0)  =  ±(n-l)!. 
Substituting  in  Maclaurin's  series, 

loge(l+2/)^,-J-f|-J  +  f-J  +  ^-... 
^     y"-1     Tyn        1 


(n-l)l     »!(1  +  fy)n 

By  Cauchy's  test  this  series  is  convergent  for  values  of  y 
numerically  less  than  unity.  Hence  this  series  may  be  used 
to  calculate  the  Napierian  logarithms  of  numbers  from  0  to  2. 


-1  o    +1 


Fro.  72. 


Taking    y=.5    and   n  =  13,    log,  1.5  =  .40546914   with   an 
error  between  .000000053  and  .0000093. 


EXPANSIONS  181 

The  region  of  convergence  of  this  expansion  of  loge  (1  +  y) 
is  indicated  graphically  by  the  heavy  line  of  the  figure. 

From  this  expansion  of  loge  (1  +  y)  an  expansion  of 
loge  (1  +  z)  with  an  enlarged  region  of  convergence  is  obtained 
by  the  following  analysis  : 

for  -  1  <  y  <  1. 

ge{       V)~    f     2      3     4     5      6      7      8 

for  -  1  <  y  <  1. 

By  subtraction, 
\oge(l  +  y)-log,(l-y)  =  log,^±^  =  2(y+t  +  t  +  l  +  ...\ 

for  —  1  <  y  <  1.     Substituting 

y  =  ^—,  \±lwmi±l%  and  when  -  1<  y  <1,  *  >0. 

L  z  +  1    1  —  2/  z 

Hence  loge  (1  -f-  z)  =  logez 

Lfl/Li_j       1       ,       1       i       1       i    > 

\2z  +  l     3(2z  +  l)3^5(2z-f  l)5^7(2z+l)7^     J 

for  2  >  0 ;   a  convenient  formula  for  the  calculation  of  the 
Napierian  logarithms  of  numbers. 
By  Maclaurin's  series 

loga(l+2/)^logaeJ2/-|2-f|-^  +  |5-|+-..}, 

that  is,  loga  (1  +  y)  =  loga  e  •  log,  (1  +  y)-     Placing  1  +  y  =  o, 
logaa  =  logae.logea, 

whence  logae  =- and  loga(l+y)  =  - loge(l  +  ?/). 

logea  logea 

The  factor ,  by  which   the   Napierian    logarithm   of  a 

loge  a 

number  must  be  multiplied  to  obtain  the  logarithm  of  the  same 


182       DIFFERENTIAL   AND  INTEGRAL   CALCULUS 

number  in  the  system  whose  base  is  a,  is  called  the  modulus  of 
the  system   of  logarithms  whose  base  is  a.     In  the  common 

system  a  =  10,  and  — - —  =  .43429448. 
loge10 

Hence  log10  (1  +  y)  =  .43429448  log,  (1  +  y). 

Example  II.  —  Expand  loge  y. 

Here  f(y)  =  loge  y,  f  (y)  =  *  f"  (y)  =  -  i,  f">  (y)  =  %  -  , .  ■ 

y  y  y 

f'iy)=±  ^^.    Hence,/(0)=-oc,  f{0)=a>,  /»(0)«-cd> 

/'"(0)  =  oo,  •••,  /H(0)  =  ±co.  The  function  logey  cannot  be 
expanded  by  Maclaurin's  series  into  a  power  series  in  y. 
However,  writing  log  y  =  log  \  1  -f-  (y  —  1)  | ,  Exainj)le  I.  gives 

a  power  series  in  y  —  1  convergent  when  0  <  ?/  <  2. 

Example  III.  —  Expand  (x  -f  ?/)™,  where  m  represents  any 
finite  number,  positive  or  negative,  integral  or  fractional. 
This  expansion  is  effected  by  Taylor's  series. 

Here      f(x)  =  xm,  f  (a?)  =  mxm~\  f"  (x)  =  m  (m  -  1)  xm~'2,  •  •  • 

/"-1  (x)  =  m  (m  -  1)  •••  (m  —  n  +  2)  .Tm-"+1, 
/"  (a;)  =  m(m-l)...(m-n  +  2)  (m  —  n  + 1)  as"*"*, 
Substituting  in  Taylor's  series, 

(x  -f  y)m  =  xm+  m  •  xm~l .  2/  +  m  (m  "~  *)  a*- Y 

m(w-l)  0-2)^,3  . 
3! 
ro(m-l)  — (ro-n  +  2)        +1     +1 

(•-1)1  ' 

[  m(m  -  1)  ...  (m  -  n  +  2)  (m  -  n  + 1)^^,  +  .... 


EXPANSIONS  183 

The  ratio  of  the  nth.  term  to  the  (n  —  l)tli  term  is 
r  =  /m  +  l_iy      gince  m  ig  finite    the  factor  m  +  l_  j 

\     n  Jx  n 

approaches  unity  when  n  is  indefinitely  increased.     Hence  if 

-  <  1,  the  ratio  r  becomes  less  than  unity  when  the  series  is 
x 

sufficiently  extended,  and  the  series  is  convergent  by  Cauchy's 
test.      This   proves  the  binomial   theorem  for  all   finite  ex- 

y 

ponents,  provided  —  1  <  -  <  1. 
x 

Example  IV.  —  Expand  (2  —  x  +  y)"~*  in  ascending  powers 
of  y. 

Substituting  v  =  2  —  x,  (2  —  x  +  y)~ 2  becomes  (y  +  y)~  *.  By 
Taylor's  series, 

(v  +  y)  "*  =  v~*  +  —  (v"h  •  y  +  —  (v'h  •  ^  +  —  (v~h  £  +... 
5  v~i  _  J  v-l .  i/  + 1 W"i  .  ^2  _  _5_ „-*  .  y»  +  ... . 

whence,  (2—x+y)~'2  =  (2  —  xf*  —  £  (2  —  a?)"*  •  y 

+  f(2-a;)^.2/2-T^(2-^.^+..., 
y 
convergent  when  —  1  <  s  _    <  1. 

PROBLEMS 
Expand  and  determine  the  region  of  convergence, 

1.  sin  y.  5.  sin(x  —  y).       9.  log  (#  +  #). 

2.  cosy.  6.  cos(#  —  y).     10.  tan_1#. 

3.  sin(x  +  y).  7.  ey.  11.  sin_1#. 


4.  cos(aj  +  y).  8.  <ry.  12.  log  \x  +  VI  H-o58}. 

13.  sinh #  =  £ (ex  —  e~x).         14.  cosh#  =  A-(ex+e~*). 

15.  e^sina.  18.  tana;.  21.  (1  —  x  +  y)K 

16.  e-8in-1*.  19.  logcosx.  22.   (a2  —  f)~l. 

17.  (l-sin2a)~*.      20.  cot?/.  23.  log  (a3  —  y2). 


184       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

10° 

24.  Compute  sin  10°.     Here  x  =  • 

o»  .o 

25.  Compute  sinh  .2. 

26.  Compute  loge  2  and  log10  2.  27.  Expand  e-*2. 

Art.  68.  —  Euler's  Formulas  for  Sine  and  Cosine 

Let   the    series    e-s  1  +*  +  g  +  g+J  +  S  +  S+  '"' 

which  is  the  development  by  Maclaurin's  formula  of  ez  when 
z  is  real,  be  adopted  as  the  definition  of  ez  for  all  values  of  z, 
real  and  complex.  Placing  z=  ix,  where  i  stands  for  V—  1, 
whence  i4n  =  1,  ^4n+1  =  i,  iu+2  =  —  1,  i^+3  =  —  i  for  all  integral 
values  of  n, 

(1 )  eix  =  1  +  ix  -  —  -  i  —  +  —  +  i  —  -  —  -  i  —  +  •  •  •  • 
v  ;  2!       3!     4!       5!      6!       7! 

In  like  manner,  placing  z  =  —  ?'#, 

v  ;  2!       3!     4!       51     6!       7! 

Taking  half  the  sum  of  (1)  and  (2), 


(3)e    +e      EEl-^+^-^  +  ^-..-^cosa; 
w  9  9!       4.        fit       «f 


dividing  half  the  difference  of  (1)  and  (2)  by  2z', 

e««_  e-fa  of      aj8      a;7   .  a;9 

(4 )  =# •••  =  sm  x. 

w       2i  815!     7!     9! 

In  general,  sin  (n#)  =  i  J  efna:  -  e-'na:  J ,  cos  (nx)  =  }  [  e1*"*  +  <r*" 1  • 
2i 

These  results  are  known  as  Euler's  formulas   for  sine  and 
cosine. 

These  formulas  are  useful  in  trigonometric  transformations 


EXPANSIONS  185 

and   in  integrating   derivatives   involving  trigonometric  and 
exponential  functions. 

Since  the  i  in  u  =  ix  is  of  the  nature  of  a  constant  factor, 
du_  . 
dx 

Example   I.  —  Find   the  value  of  cos5  a;  in  terms   of  the 
functions  of  the  multiples  of  x. 

cos5x  =  {%(eix  +  e-ix)\5 

=  ^ei5x  +  5  e*x  +  10  eix  +  10  e~ix  +  5  e"'3*  +  *-*»} 

=  tV  i  &«*"  +  e~i5x) + !(*""  +  ^,-3x) + W  +  O  J 

=  3^ {cos  (5  ic)+  5  cos  (3  x)  + 10  cos  a;}. 

Example  II.  — Integrate  -^  =  sin2ic  •  cos2  a;. 

dx 

Substituting  for  sin  x  =  —  (eix  —  e_£z)  for  cos  x  =  %(eix  +  e_fa), 

*  =  -  JL  (««-  +  at*"  -  2)  =  -  I  cos  (4  x)  +  L    Hence 
dx 

y  =  -Tfesin(4»)+|a>  +  a 

Example  III.  —  Integrate  -^  =  e-p*  sin  (pt),  where  P  and  p 

ctz 

are  constants. 

Substituting  sin (pt)=—(eipt  —  e~ipt), 
Z  i 

dt=2i16  *  s' 


hence 


9      2i\-P+ip  P+ip  J 


186       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

PROBLEMS 

Find  in  terms  of  the  functions  of  the  multiples  of  x, 
1.    sin4 a;.  2.    sin7  a;.  3.    sin2#  •  cos2  (2  a;). 

Integrate,  4.  -^=sin3#.  5.  -^=sin4#. 

dx  dx 

6.    ^=cos2x.  7.  ^=sin6#.  8.  -^  ==  e2*  •  sin2  a;. 

dx  dx  dx 

9.    Show  that  eix  =  cos  x  -f-  i  sin  x,  e~ix  =  cos  x  —  i  sin  x. 

Art.  69.  —  Differentiation  and  Integration  of  Power 

Series 
Consider  the  power  series, 

(1)    f(x)  =  a0+a1'X  +  a2-x2-\ 

an  •  xn  +  an+1  •  xn+l  +  aH+2  •  xn+2  H , 

and  denote  the  sum  of  the  first  n  terms  by  sn  (x),  the  sum  of 
the  remaining  terms  by  rn  (x). 

Absolute  Convergence.  —  Denote  the  numerical   or  absolute 
value  of  any  quantity  z  by  the  notation  \z\.     So  that  |  —  5  |  =  5, 

Suppose  in  the  power  series  (1),  denoting  \an\  by  An,  that 
AnX0H  <  M,  where  M  is  a  finite  quantity,  for  all  values  of  n. 

:zXo O +Xo_ 

Fig.  78. 

M 

Take  |  x  \  =  X  <  X0 ;    then,    since   An  <  — -  by  hypothesis, 

A  +  Ai  +  AlH-iJH- 

xj    \xj  T        \xj  T 

$ince  by  hypothesis  Jf  is  finite  and  •—  <  1,  by  Cauchy's  test 


EXPANSIONS  187 

the  right-hand  member  of  the  inequality,  and  consequently  the 
left-hand  member,  is  a  convergent  series.  That  is,  the  sum  of 
the  absolute  values  of  the  terms  of  the  original  power  series 
is  convergent  for  |  x  |  <  X0  if  AnX0n  <  M.  This  is  expressed 
by  saying  that  the  given  power  series  is  absolutely  convergent 
within  the  region  —  X0  <  x  <  X0. 

Example.  —  In  the  expansion, 

(1  -x^-i  =  1  +J  +  K  +  A*6  +  tsA*8  +  t¥i*,°+  -, 

when  x  =  1,  each  term  is  less  than  2.  Hence,  the  series  is 
absolutely  convergent  for  —  1  <  x  <  1. 

Uniform  Convergence.  —  Writing 

i0+i1i+Ai2+-A^n+A+i^+,4--W+^(i), 

rm(x)  >  Rn  (x)  <  ^(§J+  M ( JY+1+  -  for  |  ^  |  =  X  <  X0. 

— (SMD"+Jf(r+"-*(i)nr 

Now,  n  may  be  taken  so  large  that,  for  all  values  of  X  <  X0, 


Jf f  —  J —  for  the  same  value  of  n  becomes  less  than  c, 

however  small  c  may  be  assumed.  Consequently,  since 
rn(x)  >  Rn(X),  for  all  values  of  x  between  —  X0  and  -f-  XQ 
the  expression  rM  (x)  can  be  made  less  than  c  for  one  and  the 
same  value  of  n.  This  fact  is  expressed  by  saying  that 
the  power  series  is  uniformly  convergent  in  the  region 
-  X0  <  x  <  X0. 

Example. — The  power  series  l-h^+^+^H af+of~1H 

is  absolutely  convergent  for  |  x  \  <  1.     Assuming  c  =  .000001, 


188       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

determine  n  so  that  for  this  value  of  n  rn  (x)  <  c  for  all  values 
of  x  between  —  \  and  -f  \. 

Since  rn  = ,  by  the  conditions  of  the  problem  rn<(|)n_1. 

J.  —  X 

The  conditions  of  the  problem  are  satisfied  when  (£)n~1=. 000001, 
that  is,  when  n  =  22. 

Continuity.  —  Denote  by  x  and  x0  any  two  quantities  numeri- 
cally less  than  X0.     Since  f(x)  —  sn  (x)  +  rn  (x), 

f(x)  -f(x0)  =  \sn(x)  -  sn(x0)}  +  \rn(x)  -  rn(x0)\. 

Since  n,  however  large,  is  supposed  to  be  finite,  x  may  be 
taken  sufficiently  near  x0  to  make  |  sn  (x)  —  sn  (x0)  I  <  e,  however 
small  c  may  be  assumed ;  and  since  by  hypothesis  x  and  x0 
are  within  the  region  of  uniform  convergence  of  the  power 
series,  n  may  be  taken  so  large  that  |  rn  (x)  |  and  |  rn  (a?0)  |  each 
become  less  than  e.  Consequently  |  f(x)  —  f(x0)  |  <  3  c,  and 
the  function  defined  by  the  power  series  is  continuous  in  the 
region  of  uniform  convergence. 

Integration.  —  To  show  that  between  limits  within  the  region 
of  uniform  convergence  the  limit  of  the  sum  of  the  integrals  of 
the  terms  of  a  power  series  is  the  integral  of  the  limit  of  the 
power  series,  write 

Jf(x)  dx  =  I   a0  •  dx  +  J    ax  •  x  dx  -f  |    a2  •  x*dx  -\ 

+J    rn(x).dx, 

where  a  and  /3  lie  within  the  region  of  uniform  convergence  of 

f(x)  =  a0  +  «i  •  a  +  «2  •  &  +  «3  •  '&  H 

Now    J    rn  (a;)  •  dx  <  e  |    dx  =  e(fi  —  a),  where  c  is  a  quantity 

which  approaches  zero  as  n  approaches  infinity.  This  proves 
the  proposition. 


EXPANSIONS  189 

The  series 

f(x)  =<h-$ (x)  +  a1-il/(x).<l>(x)+a2'*lt  (a?)  •  \_<f>  (x)']2  + ... 

may  be  integrated  term  by  term  between  the  limits  x  =  a  and 
x  —  fi  provided  a  and  /3  lie  within  the  region  of  uniform  con- 
vergence of  the  series  a0  +  aL  •  <f>  (x)  +  a2  •  [<£  (#)]2  +  •  •  •  and  if/  (a?) 
is  finite  from  x  =  a  to  x  =  ft. 

Example  I.  —  Expand  tan-1  a;  into  a  power  series  by  inte- 
gration. 

—  tan-1*  =  — !— ^ ■  =  1  -  x2  -f-  a4  -  a6  +  x*  -  x10  +  .... 
cfo  1  +  a2 

a  power  series  uniformly  convergent  for  |  x  |  <  1.  Hence  term 
by  term  integration  gives  a  valid  result,  and 

/y»3  />i5  a»7  /x»9  /\i  1 1 

taa-»»  =  *_|+|-|+|_g+...  for  |*|<i. 

From  this  expansion  is  obtained  Euler's  series  for  the  calcu- 
lation of ».     Writing 

tan  u  =  £  and  tan  v  =  J,  tan  (w  -f  v)  =  1  =  tan^. 
Hence        j  =  ti  -f-  v  =  tan  _1  £  -f-  tan-1  £, 
and  |=g-^+^~...J+(s_§__  +  _-...) 

\2     3)     3^2"     3V     SV2*     37 
Example  II.  —  Find  the  value  of  t  when 

V2^^«   ^/(h-y)(2ry-y2) 

Here  £  is  the  time  of  vibration  of  a  simple  pendulum  of  length  r, 
the  bob  starting  at  a  distance  /i  above  the  horizontal  through 
its  lowest  position. 


190       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Substituting  y  =  h-  sin2  0, 

2  7i  •  sin  0  •  cos  6  •  d$ 


=    4r      Cl 


V2  g  J»    Vh  •  cos  0  ■  sin  0  V2  rfc  -  /i2  sin2  0 


2^Sm  *      <W 


#V*?Tr-# < 


+I.I.I.(^.^+...J» 

WiiKlJi+(H)'(r 


-'iiD'^y- 


•••} 


If  7i  is  small  compared  with  2r,  t  =  2wd-  is  an  approxima- 
tion sufficiently  accurate  for  most  purposes. 

Differentiation.  —  To  find  under  what  conditions  the  sum  of 
the  derivatives  of  the  terms  of  a  power  series, 

f(x)  =  a0  +  ax  •  x  +  a2  •  x2  +  a3  •  x3  +  •••  an  •  o?n  H , 

uniformly  convergent  for  |  a;  |  <  X0,  is  the  derivative  of  the 
function  defined  by  the  power  series,  write 

<f>  (x)  =  ax  +  2 a2  •  x  -f-  •••  ?i  •  an  •  #n-14-  (?i  -f  1)  •  aM+1  •  xn  +  •••. 

By  hypothesis  the  series  defining  f(x)  is  uniformly  convergent 
for  |  x  |  <  X0.  If  from  and  after  some  fixed  term  the  ratio  of 
the  corresponding  terms  of  <f>(x)  and  f(x)  is  not  greater  than 
unity,  <f>  (x)  is  uniformly  convergent  when  f(x)  is  uniformly 


EXPANSIONS  1U1 

convergent.      The  ratio  of  the  (n  +  l)th  terms  of  </>  (x)  and 

f(x)  is  n[-*\  — ,  where  x$  and  xf  denote  respectively  the 
\xj  x+ 

variables  of  the  <f>  and  /  series.     Hence  <£  (x)  is  uniformly  con- 


vergent when 


XfJ    Xq 


•<\/> 


>\x<t>\.     If 


>  1,  that  is,  when  w 

aty  is  taken  less  than  xf,  limit  n(  —  J  =0  when  limit  n  =  go. 

Hence  </>  (#)  is  uniformly  convergent  for  |  x  |  <  X,  when  X 
lies  within  the  region  of  uniform  convergence  of  f(x).  For 
these  values  of  x,  <£  (cc)  may  be  integrated  term  by  term,  and 
f(x)  —  f  </>  (as)  dx  =  a0  +  %  •  a  +  a2  •  a2  +  a3  •  a3  +  •  •  •,  when  a 
and  /?  lie  within  the  region  of  uniform  convergence.  Differ- 
entiating this  result,  f'(x)  —  <£  (x)  =  0 ;  hence 

f'(x)  =  a1  +  2a2-x-\ (-  n  •  aB  •  »M_1  H . 

That  is,  a  uniformly  convergent  power  series  may  be  differ- 
entiated term  by  term  as  long  as  x  is  within  the  region  of 
uniform  convergence  of  the  power  series. 

Example.  —  The  expansion 

(1)  (l-rf)-*-al+i^+i^+i-»-t^+i-f-|-i^+- 

is  uniformly  convergent  for  |  x  |  <  1.     Obtain  the  expansion  of 
(1  -  x?)~%  and  (1  -  x2)"^  by  differentiation. 
Differentiating  (1)  and  dividing  by  x, 

(2)  (1-^-1  =  14-1^  +  -^  x*  +  ffa6+...; 
differentiating  (2)  and  dividing  by  3  x, 

(3)  (1  -  x2)-*  =  1  + 1  a2  +  ^5-  x4  +  .-. 

PROBLEMS 

Expand  by  integration, 

1.    log(l  +  #).  2.    log(l  —  x).  3.    sin-1^. 


192        DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

From  the  expansion  of  (1  —  a?)-1  obtain  by  differentiation  the 
expansion  of, 

4.    (l-x)~2.  5.    (l-x)~3.  6.    (l-x)-\ 

7.  Find  the  length  of  the  ellipse  x  =  a  cos  <f>,  y  =  b  sin  <£. 
If  the  arc  is  measured  from  the  end  of  the  major  axis, 

s  =  a  I     VI  —  e2  sin2  <j>  d<f>, 

where  e  is  the  eccentricity  of  the  ellipse. 

The  entire  length  is  4a  J    VI  —  e2  sin2 <j>d<f>. 

8.  The  discharge  of  water  per  second  through  a  circular 
orifice  of  radius  r,  the  plane  of  the  orifice  being  vertical, 
when  h  is  the  head  of  water  on  the  center  of  the  orifice,  is 
Q  =  2  C+r  V^^?  -\/2g(h-y)  dy.     Find  Q. 

X  +  h 
e~x  •  dx.     This   is 

called  the  probability  integral. 

XI  /»*   g— 3 
c?ic.     Expand  ex  and  e-*  separately, 

and  express  in  the  form  of  an  infinite  series. 

x 

Art.  70. — Expansion  of  n1=f(x-{-h,  y  +  k) 

Let  u=f(x,  y)  denote  a  continuous  function  of  two  inde- 
pendent variables.  Denote  by  u1=f(x-{-  h,  y  +  k)  the  value 
of  u  when  x  and  y  are  increased  by  h  and  k  respectively.  ux 
is  to  be  expanded  into  an  equivalent  series  in  the  ascending 
powers  of  h  and  k. 

Denoting  by  u0  the  value  of  u  when  x  is  increased  by  h  and 

y  remains  unchanged,  by  Taylor's  series, 

r/     ,   7      v  .  du   ,    .  d2u    h2  .  d3u  h3   ,  d*u   hA   , 

o     j\    t    ,*;         -r^      ^do2  2!      da3  3!      dz4  4! 


EXPANSIONS 


193 


Now  Ui  is  the  value  of  u0  when  y  is  increased  by  k,  x  remain- 
ing unchanged.     Hence  Vq  =f(x  +  h,  y  +  k) 


«o  +  ^-°-fc  + 


sw  + 


5«o 

da 


By2 


7     I    ^ 


dy 


d2u 
dxdy 

dy2 


2r 


ax 


ft*        &u 

2!        dar1 


/jft-f 


5! 
2! 


a»w 


da;-d?/ 
d3w 


da;d#2 


+   H 


A?  d4ll0 

3!  +    dif 

ft?  &u 

3!  da;4 


4! 

ft* 

4! 


dar%    3! 
3*u     ft2&2 


2! 

2!      da;% 


V    3! 


£  + 


d4u 


2    2! 


tod?/3    3! 
dtf      *4! 


+ 


+  •■ 


+ 


+ 


+  •■ 


For  example,  in  the  sphere  x2  +  y2  +  z2  =  25,  at  the  point 

(0  +  /i,   4  +  &,    «),    ^  =  3-|A;-i/i2-iA;2 .      If    h  =  .5 

and  ft=.l,  z  =  2.78. 


CHAPTER    XI 

APPLICATIONS   OF  TAYLOR'S   SEKIES 
Art.  71.  —  Maxima  and  Minima  by  Expansion 

If  the  function  y  =  f(x)  is  continuous  in  the  neighborhood 
of  (x0,  2/0),  and  yl  denotes  the  value  of  y  corresponding  to 
x  =  x0±  h,  by  Taylor's  series, 

(1)  2/i  -  ft  =  /'(*)  (±K)+  f'%)  ^f  +  /'"  («.)  ^f  +■- 

If  h  approaches  zero,  yx  —  ?/0  approaches  the  term  of  the 
right-hand  number  of  (1)  which  contains  the  lowest  power  of 
h.  Hence,  if  f'(x0)  =£0,  y1  —  y0  changes  sign  with  h,  and  y0  is 
neither  a  maximum  nor  a  minimum ;  if  f'(x)  =  0  and  f"(x0) 
is  negative,  y  —  y0  is  negative  for  +  h  and  —  h,  and  y0  is  a 
maximum  ;  if  /'(#0)  =  0  and  f"(x0)  is  positive,  yx  —yQ  is  posi- 
tive for  +  h  and  —  7i,  and  y0  is  a  maximum ;  if  /'(#0)  =  0, 
f"(x0)  =  0  and  f'"(x0)  ^  0,  yl  —  y0  changes  sign  with  h,  and 
y0  is  neither  a  maximum  nor  a  minimum. 

In  general,  if  the  first  derivative  in  the  expansion  (1)  not  to 
vanish  is  of  an  odd  order,  the  function  is  neither  at  a  maxi- 
mum nor  at  a  minimum ;  if  the  first  derivative  not  to  vanish 
is  of  an  even  order,  the  function  is  at  a  maximum  if  this 
derivative  is  negative,  at  a  minimum  if  this  derivative  is  posi- 
tive.    This  agrees  with  the  results  of  Art.  24. 

194 


APPLICATIONS  OF  TAYLOR'S  SERIES  195 

If  the  function  u  =  F(x,  y)  is  continuous  in  the  neighborhood 
of  (x0,  y0),  and  u0  denotes  F(x0,  y0),  ux  denotes  F(xQ  ±  h,  y0  ±  k)  j 
by  Taylor's  series, 

(i)„1-«0=|(;0+g(fc) 

d2F    ,9  ,   o   d2F       77    ;  d2F 


2\dx2  dx0dy0  dy02        j 


1{& 

3!{dz03 


3!  (day5  day%o  dx0d2y02  dy 


,      d^  a^  dip      ,     ,   of  dF  d2F       , 

where  — ,  — ,  — - ,  •••  denote  — ,  — ,  — ,  •••  when  x=x0,  y=yQ. 
dxQ  dy0  oxj  ox   dy    dar 

If  h  and  k  approach  zero,  wx  —  u0  approaches  the  sum  of  the 
terms  of  the  right-hand  member  of  (1)  which  are  of  the  lowest 

dimensions  in  h  and  k.     If  either  or  both  — ,  — -  are  different 

dx0  dy0 

from  zero,  ux  —  u0  has  different  signs  for  different  values  of 

h  and  k,  and  u0  is  neither  a  maximum  nor  a  minimum.     If 

dF     A    dF     A        ,  d2^      d2^       62jP  .        , 

—  =  0,  —  =  0,  and  — ,    ,   —  are  not  each  zero,  u0  is 

da?0  dy0  dx02     dx0dy0     dy02 


ri    hi  ci    ri  ci    ti 

a,  maximum  if  — Ji2  +  2 hk  H -k2  is  negative  for  all 


d2^72  ,  o   d2^    77    ,  d2^ 
— ,  h2  +  2 hk  H 

oa©         ,  &»0d?/o      ,  W 

signs  of  h  and  fe,  a  minimum  if  this  expansion  is  positive  for 

all  signs  of  h  and  k. 


Writing  g.»+t"    .»  +  ".» 

day*        da;„d?/0  /       (day*    d?/02      \dx()2dy02J  \ 

dxf 
it  is  seen  that  iq  —  w0  is  negative  for  all  values  of  h  and  k,  and 
ponsequently  u0  is  a  minimum,  when  —  and  —a  are  both 


196       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


negative,  and  - — ■ >  f -—  j :  ux  —  w0  is  positive,  and  wc 

d#o     dyo        \dx0dy0J 


d2F 


d2F 


a    minimum,   when    —    and    — -    are    both    positive,    and 
dxQ2  dyQ2 


d2F   d2F      /  d2F  V 
dx02    %o2       \dx0dy0J 


Example.  —  Find  the  dimensions  of  the  rectangular  parallel- 
opiped  of  maximum  volume,  sides  parallel  to  the  coordinate 

axes,  that  can  be  inscribed  in  the  ellipsoid  —  -f-  ^-  +  —  =  1. 

a2     62     c2 

The  volume  of  the  parallelopiped  is 

Z 


Fig.  74. 


V=Sxyz  =  Sexy 


\       a2     b2J 


If    F  is  a  maximum,    Vx  =  x?y2  —  ^-| 1-  is  a  maximum, 

a2         &2 

and  vice  versa.     Forming  the  partial  derivatives  of  Vi, 
^^_2     o_4^2     2xy* 


dx 
32/ 


2a^2 


2afy 


62  ' 


^^,00     ,12  «y     2.y* 
a»2         *  a2  &2' 


APPLICATIONS  OF  TAYLOR'S  SERIES  197 

d2Fi      o   i     2 a4      12 aY 


2^-^.- 


a/  a2  62    ' 

d2Fi       ,  8a^y     8JBV8 

to  %  *        a2  62 

The  conditions  ^  =  0,    ^J  =  0    make   x=-^-,    y  =  -^-. 

These  values  of  oj  and  y  make 

52F1==_86_2    ^F1=_8^     d2V1  =      4afr 
6^  9  '    fy2  9  '   tocty  9  ' 

Fi  is  a  maximum.     Hence  the  dimensions  of  the  maximum 
parallelopiped  are  — — ,  — ,   — ^. 

PROBLEMS 

1.  A  box  with  open  top  in  the  form  of  a  rectangular  paral- 
lelopiped contains  108  cubic  inches.  What  must  be  its  dimen- 
sions to  require  the  least  material  in  construction  ? 

2.  Find  the  point  (x,  y,  z)  the  sum  of  the  squares  of  whose 
distances  from  (a1}  bX}  Ci),  (a2,  b2,  c2),  (as,  b3,  c3)  is  a  minimum. 

3.  The  sum  of  the  three  dimensions  of  a  rectangular 
parallelopiped  is  a.  Find  the  dimensions  when  the  volume  is 
a  maximum. 

4.  Find  the  dimensions  of  the  cistern  of  maximum  capacity 
that  can  be  built  out  of  3000  square  feet  of  sheet  iron,  the 
cistern  being  of  the  form  of  a  rectangular  parallelopiped  and 
without  lid. 

5.  The  sum  of  the  three  dimensions  of  a  rectangular 
parallelopiped  is  a.  Find  the  dimensions  when  the  surface  is 
a  maximum. 


198       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Art.  72. —  Contact  of  Plane  Curves 

Plot  the  curves  representing  the  equations  Y=  F(X)  and 
y  =f(x)  to  the  same  coordinate  axes,  and  denote  by  5^  and  y0 


the  values  of  Y  and  y  corresponding  to  X: 


:  Xi),  x  - 


■■x0,  by  Y1  and  yx 


the  values  of  Y  and  y  corresponding  to  X=x0±h,  x  =  x0±h. 
By  Taylor's  series, 


Ti  =  F0 


dY0(    h.     d2Y0(±h)2    <PY0(±h)>    d*Y0(±hy 
dX0K      J     dX2    2!       dX03    3!        dX0*    4! 


Vi  =  y0  +  P(±h)  + 


d\(±hy    d*!h(±hf    d%(±hy 


Hence 


dx0 


dx02    2!     +dx03    31 


+ 


+ 


ri-2/i  =  (iro-2/o)  + 


p\±h) 


+ 


dY* 

dXQ     dx0J 

d*Y0     d%\(±hf 
dxf)     3! 


dx0*     4 ! 
#K     d*y0\(±hy 


dxQ2J    2! 


dX0* 


The  curves   Y=  F(X),  y=f(x)  have  a  common  point  if 
Y0  =  y0  when  X0=av     The  difference  between  the  ordinates 


'  Y 


Fio.  75. 


corresponding  to  x0  ±  h  when  h  approaches  zero  is  of  the  first 

dY      dv 
degree  in  h  if  — 5  ^  _^P .  0£  ^g  secon(j  degree  in  h  and  the 

dXQ     d  x0 
curves  are  said  to  have  contact  of  the  first  order,  if 


d  Y0  _  dy0      A  d2  Yn      d2y0 . 


dX0     dx0 


dX02 


dx() 


APPLICATIONS  OF  TAYLOR'S  SEMES  199 

of  the  third  degree  in  h  and  the  curves  are  said  to  have  con- 
tact of  the  second  order  if  ^  =  *fo  d?Y«  =  d%  and 
dsYo      d3yo  dXo       dxo     dXd        d®o 

]X3^ ~M*'     "^n  Seneralj  the  difference  between  the  ordinates 

is  of  the  (n  +  l)th  degree  in  h  and  the  curves  are  said  to  have 
contact  of  the  ?ith  order  when  the  first  pair  of  corresponding 
derivatives  not  equal  are  of  order  n  -f- 1. 

If  the  contact  of  the  two  curves  Y=F(X)  and  y=f(x)  at 
(#0,  y0)  is  of  an  even  order  2  m, 


Y-v  =(d2m+1YQ     ^tm+\\(±h)am+1 
1     Ul      \dX02m+1     d^+V(2m  +  l) 


Hence  Yx  —  y1  changes  sign  with  h  and  the  curves  intersect. 
If  the  contact  is  of  an  odd  order,  the  curves  do  not  intersect. 

Suppose  the  equation  y=f(x)  to  be  completely  determined, 
while  the  equation  Y=F(X)  involves  arbitrary  constants, 
that  is,  parameters.  The  condition  necessary  for  the  inter- 
section of  the  curves  represented  by  the  equations  Y—  F(X) 
and  y  =/(#)  when  X0  =  x0,  namely  Y0  =  yQ,  determines  one  of 
the  parameters  of   Y—  F(X)-,   the  conditions  for  contact  of 

the    first    order   when    X0  =  x0,    namely    Y0  =  y0,   — — ?  =  -^, 

dX0      (xXq 

determine  two  parameters  of   Y=F(X)-,   the  conditions  for 

contact  of  the  second  order  YQ  =  y0,  — — °  =  -%   — — f  =  — ^ 

dX0      dx0    dXo       dx<f 

determine  three  parameters  of  Y=  F(X).  In  general,  con- 
tact of  the  nth  order  determines  n+1  parameters  of  Y=F(X). 
It  is  also  evident  that  the  highest  order  of  contact  Y=  F(X) 
can  in  general  have  with  another  curve  y=f(x)  is  one  less 
than  the  number  of  parameters  of  7=  F(X). 

Example  I.  —  Determine  the  order  of  contact  of  4  y  =  x2  —  4 
andar2  +  2/2-2i/  =  3. 


200        DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  common  point  is  (0,  —  1).  For  this  point  the  first  equa- 
boa  gives  ^  =  0,  gmf  ^=0,  ^=0,  ...,  the  second 

equation  2  =  0,   g  =  |,    g=0,    g  — |     Hence  there 

is  contact  of  the  third  order. 

Example  II. — The  equation  y=f(x)  represents  a  fixed 
curve,  (X  —  m)2  +  (  Y  —  n)2  =  R2  is  the  equation  of  any  circle. 
The  parameters  m,  n,  R  are  to  be  so  determined  that  the  circle 
has  contact  of  the  second  order  with  y  =  f(x)  at  the  point 

0»o,  2/o). 

The  problem  requires  that 

m\     Y  =  v      dYo  =  d]k     d2Y0  _  <$% 
w        °      *•    dX0     dx0'    dX02     dx£ 

when  X0  =  x0.     Differentiating  the  equation  of  the  circle  twice 
in  succession, 

(X-  m)2  +  (F-  n)2  =  R2,    (X-  m)  +  (F-  n)g  =  0, 

By  the  conditions  (1),  these  equations  become 

(a>0-m)2  +  G/0-n)2  =  i22,     (a*  -  m)  +  (y0  -  n)  *  =  0, 

&Xq 


Whence 


l+^+(2/o-»)^2=0. 


&yi 


n      I        dab*/  V       dxo2Jdx0  ,  rifo02 

d%  d%  d2yQ 

dx02  d.r{)2  dx02 


APPLICATIONS  OF  TAYLOR'S  SERIES  201 

This  circle  is  called  the  osculating  circle  at  the  point  (x0,  y0) 
of  the  curve  y  =  f(x),  and  by  comparison  with  the  results  of 
Art.  44,  this  circle  is  seen  to  be  identical  with  the  circle  of 
curvature. 

PROBLEMS 
Determine  the  order  of  contact  of 

1.  yl  —  ^x  and  x2  -\-y2  =  4:x. 

2.  x2y  -f  y  —  x  =  0  and  y=0. 

3.  —  +  ?/2  =  1  and  x2  +  y2 +6y  -  7  =0. 

4.  Show  that  at  a  point  of  inflection  the  tangent  y  =  mx  -f-  n 
has  contact  of  the  second  order  with  y  =f(x). 

5.  Show  that  at  a  point  of  maximum  or  minimum  curvature 
of  y  =f(x)  the  osculating  circle  has  contact  of  the  third  order. 

Art.  73.  —  Singular  Points  of  Plane  Curves 

Let  (xq,  ?/o)  be  any  point  of  the  plane  curve  F(x,  y)  =  0, 
(x0  +  h,  y0  +  k)  any  other  point.     By  Taylor's  series 

ox0  oy0 

\dx02  dx0dy0  dy0       J 

Denoting  the  point  (Xq  +  H,  y0  +  fy  by  (a?,  y),  this  series 
becomes 


(x-x0)2  +  21^(x-x0)(y-y0)+^r2(y-y»y 
[  dx0*  dx0  dy0  dy0 


+  i 

+    ...: 


202       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


If  either  or  both  — ,  —  differ  from  zero,  when  (x,  y)  in- 

dx0    dy0 


definitely  approaches  (xQ,  y0),  (1)  approaches 
dF  ,  N  .    dF,  v     A 

the  tangent  to  F(x,  y)=0  at  (x0,  y0). 

If =  0  and  - —  =  0,  (1)  approaches 

dx0  dy0 

This  equation  is  homogeneous  of  the  second  degree  in  (x— *r0) 
and  (y—yo),  and  therefore  represents  two  straight  lines  through 
(#<»>  2/o)-     This  means  that  at  the  point  (x0,  y0)  of  the  curve 


Fig.  76. 


F(x,  2/)=0  two  tangents  can  be  drawn  to  the  curve.  If  the 
curve  stops  at  (x0,  y0)  and  the  tangents  are  real  and  distinct, 
the  curve  is  said  to  have  a  salient  point  at  (x0,  y0) ;  if  the  tan- 
gents are  real  and  coincident,  the  curve  is  said  to  have  a  cusp 


APPLICATIONS  OF  TAYLOR  S  SERIES 


203 


at  (#0,  y0),  of  the  first  species  when  the  two  branches  of  the 
curve  lie  on  different  sides  of  the  common  tangent,  of  the 
second  species  when  both  curves  lie  on  the  same  side  of 
the  common  tangent. 

If  the  curve  does  not  stop  at  (x0,  y0)  and  the  tangents  are 
real  and  distinct,  (x0,  y0)  is  a  point  where  the  curve  crosses 
itself,  called  a  node. 

If  the  tangents  are  real  and  coincident,  two  branches  of  the 
curve  are  tangent  to  each  other  at  (x0,  y0),  and  (x0,  y0)  is  called 
a  tac-node. 

If  the  tangents  at  (x0,  y0)  are  imaginary,  (x0,  y0)  is  an  isolated 
point,  called  a  conjugate  point  of  the  curve. 

An  ordinary  or  regular  point  of  a  curve  is  a  point  (x0,  y0)  for 

which  —  and  —  are  not  both  zero;    all   other  points   are 
dx  dy  ,2 

singular.     The  points  of  inflection  of  a  curve,  where  — \  =  0, 

dx2 
are  also  classed  as  singular  points. 

Example  I.  —  Examine  x*  —  3  xy  -f-  y3  =  0  for  singular 
points.     Here 


dF=3x*-3y,  ^=-3x+3y*f  ^=6x, 


dx 


dy 


oar 


dy2  dxdy 


The  conditions  — -  =0,  — =0 
dx  dy 

determine  the  singular  point 

(0,  0).     For  the  point  (0,  0), 

<PF     «    d2F     «     d2F 

dx2  = 


0,^=0, 

dy2  dx  dy 


=  -3. 


Hence  the  two  tangents  at 
(0,  0)  are  x  =  0  and  y  =  0, 
the  coordinate  axes.  Plotting 
the  curve  in  the  neighborhood 
of  (0,  0),  it  is  seen  that  the 
origin  is  a  node. 


Fig.  77. 


204        DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Example   II.  —  Examine    y2  —  x3  +  2  x2  =  0    for    singular 
points.     Here 

dF  o    2  ,   a       dF     o      &F  a      ,   a    &F     o     ^       A 

—  =  _  3  #2  _|_  4  #?  —  =  2 y,  — -  =  —  6  x  +  4,  — -  =  2, =  0. 

dx  dy  oar  dy2  dxdy 

The  conditions  —  =  0,  —  =  0  determine  the  singular  point 
dx  dy 

(0,  0).     For  this  point  ^?=  4,    ^=2,    ^L  =  0.     Hence 
dx2  dy2  dxdy 

the  tangents  at  (0,  0),  represented  by  4  x2  +  2  ?/2  —  0,  are 
2/  =  ±V  — 2«ic,  and  the  point  (0,0)  is  an  isolated  point  of 
the  curve. 


Example   III.  —  Examine  y2  =  x3  for  singular  points. 

jt        dF         o   ,   dF     0      d2F         r       d2F     0      d2F       rt 

Here  —  =  —  3 x2,  —  —  2y,  — -  =  —  6  cc,  — -  =  2,  =  0. 

dx  dy  dx2  dy2  dxdy 

The  conditions  —  =  0,  —  =0  determine  the  singular  point 
dx  dy 

(0,  0).     For  this  point  —  =  0,  —  =  2,  -^-  =  0.     Hence  the 

v  '    J  F         dx2        '  dy2        '  dxdy 

two  tangents  at  (0,  0),  represented  by  2  y2  =  0,  coincide  with 
the  X-axis.  Plotting  the  curve  in  the  neighborhood  of  (0,  0), 
it  is  seen  that  the  origin  is  a  cusp  of  the  first  species. 


PROBLEMS 

Examine  for  singular  points, 

1.  x?+3x2-y2+3x+4:y-l=0.       4.  y2  =  x2  +  2x*. 

2.  (y  —  x2)2  =  x\  5.  y2  =  x4  +  x5. 

3.  y2  =  -£-•  6.   (x2  +  y2)2  =  a2{x2-  y2). 

X  Li 

7.  (^  +  l)2+(a-l)3(*-2);=0. 

8.  y2  =  x4  —  af\ 


CHAPTER   XII 

ORDINARY  DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER 
Art.  74.  —  Formation  of  Differential  Equations 
An  equation  containing  ordinary  derivatives 

=  0 


F(x,  y,  % 
\         dx 


dx2 


is  called  an  ordinary  differential  equation. 
An  equation  containing  partial  derivatives 

■ujf  dz     dz     d2z     d2z\      A 

*(*  *  *>  as  w  w  wr° 

is  called  a  partial  differential  equation. 

The  order  of  a  differential  equation  is  the  order  of  the 
highest  order  derivative  occurring  in  the  equation. 

The  degree  of  a  differential  equation  is  the  greatest  expo- 
nent of  the  highest  order  derivative  when  the  exponents  of  all 
derivatives  in  the  equation  are  positive  integers. 

Example  I.  —  The  equation  (x  —  c)2  +  y2  =  \  c2  represents 
all  circles  with  center  in  the  X-axis  and  with  radius  ^  the 
abscissa  of  the   center.     Differentiating    this   equation   with 

respect  to  x,  x  —  c  +  y—  =  0.    Eliminating  c  from  this  result 

and   the   given   equation,   3y2-^-  —  2xy^-  +  4:y2  —  x2  =  0,  the 

dx2  dx 

differential  equation  of  the  given  system  of  circles.     Solving 

205 


206       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


the  differential  equation  for  -^,  -£  = — &-= 7-— £ 2-. 

dx    dx  b  y1 

Hence  for  every  point  (a?,  y)  of  the  plane  where  16  x2y2 — 48  2/4  >  0, 
the  differential  equation  determines  two  unequal  values  for 

— ;   where  16 x2y2  —  48 y*  =  0,  the  two  values  of  -f-  are  equal; 

dx  dy    dx 

where  16  x*y2  —  48  ?/4  <  0,  the  two  values  of  j?  are  imaginary. 

Geometrically  these   results  mean  that  through  points  (x,  y) 

for  which —  <  y  <  -\ two  circles  of  the  system  pass 

V3  V3 

and  their  tangents  at  (a*,  y)  have  different  directions;  through 

Y 


points  (x,  y)  for  which  y  =  ±  — -  pass  two  circles  which  have 

a  common  direction  at  this  point ;  and  through  points  (x,  y) 

x3 
for  which  y2  >  —  no  circles  of  the  system  pass. 
o 

If  x,  y,  -*-  satisfy  the  differential  equation,  (x,y,   -*-  ]  de- 
(Xx  v  dx  J 

notes  a  point  in  the  circumference  of  one  of  the  circles  of  the 
system  and  moving  along  the  circumference.     If  !x,  y,  —J 

V        •         (XX  J 

moves  along  in  obedience  to  the  differential  equation,  changing 
its  direction  continuously,  it  describes  the  circumference  on 
which  it  started. 


ORDINARY  DIFFERENTIAL   EQUATIONS  207 

If,  however,  (x,  y,  -^  )  denotes  a  point  in  either  of  the  lines 
A  dxJ 

y  =  ±  — -,  it  may  move  along  in  obedience  to  the  differential 

V3 
equation   without   a   discontinuous   change   of   direction,  and 


describe  the  straight  lines  y  =  ±  — -.     These  straight  lines  are 

V3 
the  envelopes  of  the  system  of  circles  (x  —  c)2  +  y2  =  \  c2. 

The  equation  (x  —  c)2  +  y2  =  \  c2  is  called  the  general  solu- 
tion of  the  differential  equation 

3y2%£-2xy^  +  ±y2-x2  =  0. 
dx2  dx 

The  solution  obtained  by  assigning  to  the  arbitrary  constant 
in  the  general  solution  some  particular  value  is  called  a  par- 
ticular solution  of  the  differential  equation  and  represents  a 
particular  circle  of  the  system. 

The  solution  y  =  ±  -^,  which  cannot  be  obtained  from  the 
V3 
general  solution  by  assigning  a  particular  value  to  the  arbi- 
trary constant,  is  called  a  singular  solution  of  the  differential 
equation.     If  the  differential  equation  is  written  F(x,  y,  p)=  0, 

where  p  =  -^,  the  preceding  analysis  shows  that  the  singular 

dx 
solution  is  the  p-envelope  of  the  equation. 

Example  II. — Form  the  differential  equation  of  the  sys- 
tem of  circles  x2  -f  y2  —  2  ax  —  2  by  +  c  =  0. 
Differentiating  three  terms  in  succession, 

w  u  dx  dx       ' 

w  x+{£)+yd-bd=0  or  —Hj, — =6- 

cfa2 


208       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

(3)     3^(^Y-^-^Yd^=0. 
^  '       dx\dx2J       dx?      \dxj  dar3 

Observe  that  in  Examples  I.  and  II.  the  order  of  the  differ- 
ential equation  is  the  same  as  the  number  of  arbitrary  con- 
stants in  the  general  solution.     This  is  always  the  case. 

The  solution  of  a  differential  equation  is  also  called  the 
primitive  of  the  equation.  The  differential  equation  is  ob- 
tained from  its  primitive  either  directly  by  differentiation  or 
by  the  elimination  of  constants  from  the  primitive  and  its 
derivatives.  The  process  of  finding  the  primitive  of  a  differ- 
ential equation  is  called  solving  the  equation.* 


PROBLEMS 

Form  the  differential  equations  of  the  following  primitives, 

1.    y  =  cx  +  c-<?.  2.    ;?/  =  c1ar}  +  ^- 

3.    (y -\- c)2  =  4:  ax.  4.    y  =  cxeax  +  c2e~ax. 

5.   y  =  d  cos  (ax  +  c2).  6.    y  =  c^2*  +  c2e~3x  -f-  c3ex. 

7.  Form  the  differential  equation  of  the  system  of  straight 
lines  y  =  mx  +  n. 

8.  Form  the -differential  equation  of  the  system  of  circles 
concentric  at  the  origin. 

9.  Form  the  differential  equation  of  the  system  of  parabolas 
y2  =  2px. 

*  The  mathematical  expression  of  every  physical  law  leads  to  a  differ- 
ential equation.  For  example,  the  relation  between  current  i  and  time  t 
in  a  circuit  whose  constants  are  R,  L,  C  is  expressed  by  the  second  order 

differential  equation  —  +  -  —  -f  —  =  !/'(*),  where  f(t)  is  the  elec- 

(It        L  (It      .L/C       L 
tromotive  force  expressed  as  a  function  of  time. 


ORDINARY  DIFFERENTIAL  EQUATIONS  209 


10.  Form  the  differential  equation  of  the  system  of  ellipses 

?  +  P       ' 

11.  Form  the  differential  equation  of  the  system  of  tangents 
y  =  sx  ±  Vl  -f-  s2  to  x2  +  y2  =  1  and  find  the  singular  solution 
of  the  differential  equation. 

12.  A  point  (x,  y)  generates  a  curve.  Write  the  differential 
equation  which  expresses  the  fact  that  the  angle  the  line  from 
the  origin  to  (x,  y)  makes  with  the  X-axis  is  the  supplement 
of  the  angle  the  direction  of  the  point  (x,  y)  makes  with  the 
X-axis. 

Art.  75.  —  Solution  of  First  Order  Differential 
Equations  of  First  Degree 

First  order  differential  equations  which  are  readily  solved 
occur  in  the  following  standard  forms : 

Standard  I.  —  XY&+X1Yl  =  0,  where  X,  Xr  are  fimc- 
dx 

tions  of  x  only;   Y,   Yy  functions  of  y  only.      Dividing  the 

equation  by  XYlf -H -=0,  the  variables  are  separated 

Yxdx     X 

and  each  term  may  be  integrated. 

Example.  —  Solve  (x2  -  yx2)  dl  +  y2  +  xy2  =  0.     Writing  the 
,  -   ,  dx 

equation     0  J~y h     +  x  dx  =  0,   and  integrating   term   by 

y2(l-y)         x 

term,  log ^— ! —  ==  c. 

y      %y 

Standard  II.  —  $?  =  f]^y\  where  fx (x,  y)  and  f2 (x,  y)  are 
dx     f2(x,y) 
homogeneous  functions  of  the  same  degree.     .Now,  a  homo- 
geneous function  of  degree  n, 


210       DIFFERENTIAL   AND   INTEGRAL   CALCULUS 

axn  +  bxn  ly  +  cxn~2y2  +  •••  hx2yn-2  +  fcasy*-1  +  lyn, 
may  be  written 

Hence,  ft-^fe^..^     Substituting  ?=*,  £-.+•£> 
dec     /a  (a?,  y)         \xj  &  x         dx  dx* 

the  given  equation  becomes  z-\-x—  =F(z),  whence  —  = 


dx  x      F(z)—z' 

where  the  variables  are  separated. 

Example.  —  Solve  y2  +  x2-^-  =  xy~- 
9 .         dx       ydx 


Here^  = 
dx 

xy  —  x2 ' 

X> 

y—l 

X 

the  given  equation  becomes 

z  +  zfU 
dx 

z2 
z-1' 

«  i   i-i  n      V        dy  dz 

Substituting  -  =  z,  -j-  =  z  +  Xy- ; 

X  CIX  (IX 


whence  —  =  ( 1 \dz. 


xz 
Integrating,  log  x  =  z  —  log  z  -f  log  c,  or  —  =  ez.     Substituting 

y  c 

V  I 

z  =  -,  y=c-ec. 
x 

Standard  III.  —  (ax -\- by -\- c)^- -\- Ax -\- By -\- C  —  0,  where 

dx 

a,  b,  c,  A,  B,  C  are  constants.     Substituting  (1)  x  =  x0  +  x1} 
i-ft  +  fewtog.jfc 

j  (««,+ ty0 + c) + (ob, + hy,)]^  +  (Ax, + By, + G)  +  {Ax, + By,) = 0. 

Determining  x0,  y0  so  that  ax0  +  by0  +  c  =  0,  ^4x0  +  2fy0  +  (7=0, 

whence    x0  ==     c  ~ yfl  —     a  ~  c       the    equation    becomes 

Ab-aB'  m      Ab-aB1  4 


da 
first  degree. 


{axi  +  62/j)  ^  +  Axx  -f-  5?/!  =  0,  which  is  homogeneous  of  the 
axi 


ORDINARY  DIFFERENTIAL  EQUATIONS  211 

The    substitution   (1)   is    impossible    when    Ab  —  aB  =  0. 
a     a 
In  this  case,  writing  —  =  —  =  ra,  the  given  equation  becomes 
,  a      b 

(ax  +5?/  +  c)-f  +  m (ax  +  by)  +  O=0?  where  the  variables  are 
ax 

separated  by  the  substitution  ax  -f  by  =  z. 

Example.  —  Solve  (3?/-7a;  +  7)  +  (7  2/-3a  +  3)^=0. 

dx 
Substituting  x=x0+x1)  y=y0+yi,  the  given  equation  becomes 

\(7yo-Sx0  +  3)  +  (7yl-Sxl)\^ 

+  l(3y0-7x0  +  7)  +  (3y1-7x1)l  =  0. 
Writing  7  y0  —  3  x0  +  3  =  0,  3  y0  —  7  x0  +  7  =  0,  whence  a\,  =  1, 
y0  =  0,   there   results    (7  yx  —  3  a^)  ~  +  3  yx  —  7  xx  =  0.      This 

equation  may  be  written  -Ml  = -.  which,  by  the  substi- 

a?, 

tution  —  =  z.  -~  =  z  -f  a;,— ,  becomes  —  =  =-^ =r dz.    In- 

Xi       '  dx1  dx±'  xx      7(1  —  sr) 

tegrating  by  partial  fractions, 

logs,  =  -  flog  (1  +  z)—  ^log  (1  -  z)  +  log  c. 

Substituting  z  =—,  there  results,  finally, 
x± 

(x  +  y-iy(x-y  +  iy=a 

Standard  IV. ^  +  I12/  =  I2)  where  Xl  and  X2  are  f unc- 

dx 

tions  of  x  only.     Since  this  equation  is  of  the  first  degree  in  y 

and  its  derivatives,  it   is  called   the  linear  equation  of  the 

first  order. 

Consider  the  equation  ^  +  X$  =  0.     Writing  this  in  the 
dx 


212       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

form  -^f  =  —  Xx  dx  and  integrating,  y=c  •  e~SXldx  or  y  •  e/Jl  cfa=c. 
Differentiating  this  result,  eS**(&  +  Xfl)=0. 

\CtX  J 

To  solve  -*  +  X,2/  =  X2,  multiply  both  sides  of  the  equa- 
dx 

tion  by  e^dx.  which  gives  el***($L  +  X2?A  =  e$Xldx X2.     Inte- 

r  V^x  / 

grating,     y  •  e^1*2  =  I  eJ"XldxX2cZx  +  0,  whence 


y  m  «rM-*T  fe/Xld*X2da;  +  <7~|. 


ndy=dL  f    ndy=_ 
9    dx     dxJ*    dx     1 


The  equation  -^  +  X^  =  X^71  is  reduced  to  the  linear  form 
dx  , 

by   dividing  by  yn.     This   gives    ?/""—  +  X1?/1_n=X2.      Now 

n(fy_      1      d   tl. 

Hence  the  given  equation  becomes 

±f-"  +  (1  -  n)  X,  •  2/1""  =  (1  -  n)  X,, 

which  is  linear  if  yY~n  is  considered  the  dependent  variable. 
Any  equation  of  the  form  f'(y)  -^+  Xxf{y)  =  X2  becomes 

(XX 

linear  by  the  substitution  z  =f(y),  —=f'(y)^-. 

dx  clx 

Example  I.  —  Solve  (1  +  x2)  -^  —  xy  =  a. 

dx 

Writing  the  equation  JL — y  = — - — ,  it  is  seen  to  be 

*  dx     1  +  aT      1  +  x2 

linear  with  Xx  =  -^ — ■ ,  X2  = :  a     .     Hence 
1  +  x2  1  +  x2 

Cxxdx  =  log(l  +  a2)"*  and  e^*6  =  (1  +  a2)"*. 
*  Leibnitz  seems  to  have  been  the  first  to  obtain  this  formula. 


ORDINARY  DIFFERENTIAL  EQUATIONS  213 

Substituting  in  the  formula, 

»-(l+^*f  f     «*»  m  +  0\.    Nowf     ad*     ■       ax      , 
\  J  (1  +  a2)1        J  J  (1  +  *0*     (1  +  «0* 

found  by  substituting  x  =  tan  0.     Finally  y  =  ax+C(l  +  x*)k 

Example  II.  —  Solve  ^+y  =  xy3. 
dx 

Dividing  by  y3,  y~3^-  +  y~2  =  a    Writing  y~3^-  »  A  (-  J  j^, 
otic  otic      rtic 

the  equation  becomes  — (—  2/2)  —  2y~2=  —2  a;  and 

2/-2  =e!2d*Ce-f2dx(-2x) dx  =  e2*  Ce~2x(-2x) dx. 

Integrating  by  parts, 

2/-2  =  e2*  (axr2*  +  J<r«-  +  O)  =  a;  +  J  +  Ce2*. 

Example  III.  —  Solve  3  y2  ^  -  ay3  =  a;  +  1 

eta; 

Writing  this  equation  —  (i/3)  —  a?/3  =  #  4-  1, 
eta/ 


.«-[/, 


(a  +  l)<fa;+<7 


Cfe-_*±1_1 


Standard  V.  —  Exact  equations.  A  differential  equation 
is  said  to  be  exact  when  it  can  be  obtained  directly  by  the 
differentiation  of  its  primitive.  By  Art.  33  the  equation 
Pdx  +  Qdx  =  0,  where  P  and  Q  are  functions  of  x  and  y,  is 

exact  if  —  =  —£•     The  primitive  is  found  by  the  method  of 

dy      dx  r  J 

Art.  33. 


Example.  —  Solve  x  (x2  +  3  y2)  dx  +  y  (y2  +  3  x2)  dy  =  0 

A(a^  +  3aW2)  =  6a:2/  =  f  (f  ■ 
ay  dx 

exact.     Considering  y  constant, 


—  (x3  4-  3  a*/2)  =  6  a??/  =  —  (y3  +  3  afy),  hence  the  equation  is 


214       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Differentiating  this  result  partially  with  respect  to  y,  and 
equating  to  coefficient  of  dy  in  given  equation, 

3^y  +  ^My)  =  y3  +  3xiy, 

ay 

whence  £/,  (y)  =  f  and  /,  (y)  =  \y*+C. 

dy 

The  primitive  of  the  given  equation  is  iyA+%x2y*-\-\x4+C  =  0. 

Standard  VI.  —  Integrating  factor.  If  the  equation 
Pdx  -|-  Qdy  =  0  is  not  exact,  a  factor  //.  may  be  found  for 
which  the  equation  fxPdx  +  fxQdy  =  0  is  exact,     /x  must  satisfy 

the  equation  —  (xiP)  =—  (/xQ). 

If  /x  is  a  function  of  #  only,  this  equation  becomes 

§£„    dQ+Q<!R  whence  l^  =  I/^-^A 
d#  d#         d#'  /x  da;      Q\cty       day 

and  ti  =  eJ«v^    *°;    . 

If  the  equation  Pdx  +  Qdy  =  0  is  homogeneous  of  degree 
m  and  the  integrating  factor  /x  homogeneous  of  degree  n,  the 
equation  fxPdx  +  nQdy  =  0  is  exact  and  homogeneous  of  degree 
m  +  ?i.     Hence,  by  Problem  12,  Art.  33, 

/iJRu  +  fiQy  =  (m  +  »  +  1)0. 

Since  O  is  an  arbitrary  constant,  (m  +  n  -f  1)  C  may  be  taken 

equal  to  unity  and  the  integrating  factor  u  = 

Px+Qy 

Since   -  d  (xmyn)k  =  a,-*"1-1?/* n~1(my  dx  +  nx  dy),  the  differential 

expression  xayP(mydx  +  nxdy)  is  rendered  exact  by  the  factor 
gjcm-a-iykn-p-i^  wnere  ft  js  anv  number  whatever. 


ORDINARY  DIFFERENTIAL  EQUATIONS  215 

Example   I.  —  Find  the   integrating  factor  of  the  linear 

equation  -^  -f  X-aj  =  X2. 
dx 

Supposing  the  factor  to  be  a  function  of  x  only, 
P  =  (X,y  -  X2),  Q  =  1,  and  /x  =  e  /**. 

Example   II.  —  Find  the  integrating  factor  of  the  homo- 
geneous equation  (xy  +  y2)dx  —  (x2  —  xy)dy  =  0. 

1  1 


The  factor  is  (x  = 


y?y  +  xy2  —  x?y  -f-  xy2     2  #?/ 


Hence  ^  +  *  +  ^-^  =  0 

2x     2y     2y      2y2 

is   exact.     Writing   this    equation    —  4-  ^  +  yda;~  xdy  =  Q 

a  x        y  y2 

and  integrating,  log  (a;?/)  -f  -  =  (7. 

Example  III.  —  Solve  (2  arty2  +  y)dx  —  (arty  —  3  x)dy  =  0. 
Break  up  the  equation  into  two  parts  of  the  form 

ary^my  dx  +  nx  dy),  arty(2  ydx  —  x  dy)  +  (ydx  +  3x  dy)  =  0. 
r^k-Zy-k-2  is  an  integrating  factor  of  the  first  part,  a^i-ty^i-1 
an  integrating  factor  of  the  second  part.  These  factors  are 
the  same  when  2^  —  3  =  ^  —  1,  —  k  —  2  =  3^  —  1,  whence 
k1  =  —  $  and  the  common  integrating  factor  of  both  parts  of 
the  equation  is  af  m/"*-.  Multiplying  the  equation  by  this 
factor  and  integrating,  -J  x  ^y~^  —  J  x~*y*  —  C,  whence 
4a%  =  5  +  CaM*. 

Standard   VII.  —  The   equation  fx(xy)y  dx  +  f2(xy)x  dy=0 
may  be  solved  by  the  substitution  xy  =  v,  whence 

,       xdv—vdx 
dy  =  — ■ — -9 


216       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

v  xdv d  dx 

The  equation  becomes  fi(v)-dx+fa(v) =  0,  reduc- 

.     dx  Mv)dv  ,!•!-, 

mg  to  —  =  —  where  the  variables  are  separated. 

Example.  —  Solve  (x2y2  +  xy)y  dx  +  (x2y2  —  T)x  dy  =  0. 

xdv fi)  dx 

Substituting  xy  =  v,  dy  = ,  the  equation  becomes 

x 

/  2  i     \Vj     1/2      t\Xdv  —  vdx      .      .  .  .       ,  L  dx     dv 

(v2  +  v) -dx  +  (v2  —  1) =  0,  which  reduces  to—  =  — . 

x  x  xv 

Integrating,  log  x  =  log  v  +  log  c,  x  =  ecv  and  finally  x  =  ecxy. 

In  attempting  to  solve  a  differential  equation,  determine 

what  standard  applies  and  proceed  by  the  method  of  that 

standard. 

PROBLEMS 

Solve, 

1.   x2<^--y2-xy  =  0.  2.    %L —  y  =  b. 

dx     y         y  dx     l-xy 

3.    (l-x)2ydx  +  (l+y)x2dy  =  0. 
4.   ft+y.'qf.  9.   (y-3x+S)^-=2y-x-4:. 

(XX  CLX 


5.  *22_j,=v?^?.        io.  4«jL 


a>" 


i%  = 


6.  (l+3*)-a^  =  0.  n     (x  +  yy?z  =  aa. 

dx  dx 

7.  *  +  »-«-.  12.    &+£-» 

dx  dx     x 

8.  x2ydx-(xz+f)dy  =  0.      13.    x^-y^Vx^ff 

dx 

14.  (arJ  +  l)^  +  2aw=4a2. 

die 

15.  sec2 a;  •  tan?/  •  d#  4-  sec2 y  •  tana;  •  dy  =  0. 

16.  (y-x)^  +  y  =  0. 


ORDINARY  DIFFERENTIAL  EQUATIONS  217 

17.    dy=x*f-Xy.  18.   ^-=ay2x. 

(XX  (XX 

19.  (Vxy  —  l)xdy  —  (Vary  -f  l)y  dx  =  0. 

20.  (xy-x2)^  =  y2. 

dx 

21.  xy  dy  —  y2  dx  =  (x  +  y)2  e  x  dx. 

22.  (x-y)2  +  2xyC^  =  0. 

ax 

23.  (x2  +  2xy-y2)dx=(x2-2xy-y2)dy. 

24.    1%^  +  2^-^Lo.         25.    &=^±£ 
x       y         \y       x  J  dx        2  xy 

26.  y(xy  -\-2  x^y2)  dx  +  x  (xy  —  afy2)  dy  =  0. 

27.  —  cosic-f-  vsina;  =  1. 
dx 

28.  (x?  +  3xy2)dx+(yi  +  3x2y)dy  =  0. 

29.  (^-2?/a;2)da;  +  (2a'?/2-iB3)^  =  0. 

30.    (x+y)^-+x-y=:0.        31.   ^  +  2/  cos  x  =  sin  (2  x). 
dx  dx 

32    dy  =  l  +  y  +  y8, 

da;     1  +  a;  +  ar2 
33.    a;(^-32/2)+2/(3a;2-2/2)^=0. 

34.  dy+exy  =  exy2.  36.    (1  +  a^)^+  aw  --  =  0. 
eta  c?a:  a 

35.  -2.  = —^_ !__!__.  37.    y -Z  +  by2  =  a  cos  x. 
dx     3x  +  5y  +  6  s dx       9 

38.    (1  +  xy)ydx  +  (1  —  xy)xdy  =  0. 

39.  Determine  the  curve  whose  subtangent  is  constant. 

40.  Determine  the  curve  whose  subnormal  is  constant. 

41.  Determine  the  curve  whose  subtangent  at  any  point 
equals  the  sum  of  the  coordinates  of  that  point. 


218       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


42.  The  radius  vector  cuts  a  curve  under  a  constant  angle. 
Find  the  equation  of  the  curve. 

43.  Find  the  system  of  curves  which  intersect  all  parabolas 
y2  =  2px  at  right  angles. 

Through  every  point  (x,  y)  of  the  plane  there  passes  one 
parabola,    whose    direction    at    this 

dii      T)       ii 
point  is  —  =  -  =  ^-.     For  the  curve 
dx     y     Zx 

which  cuts  this  parabola  at  (x,  y)  at 


right  angles. 


dy 
dx 


2x 

y ' 


Integrating, 


Fig.  79. 


1,  a  system  of  ellipses. 


44.  Find  the  system  of  curves  in- 
tersecting the  hyperbolas  xy  —  a2  at 
right  angles. 


Art.    76.  —  Equations   of   First   Order   and    Higher 
Degrees 


Case  I.  —  Suppose  the  equation  of  the  nth  degree 

dtf 
dx2 


dyn  ,      dyn~l  ,      dyn~2  .         .  dy2  .  dy  ,  A 


dxn  '  *  dxn~l         dx' 
to  be  resolvable  into  n  factors, 


dx 


dy 
dx 


*)(I-*)(S-*)-(2-?")=a 


The  given  equation  is  satisfied  only  when  one  of  these  factors 
vanishes.     Eepresenting  by 

/i(»>  V>  c0  =  °>  Mxt  Vi  <%>  =  0,  >~fn(x,  y,  c)  =  0 
the   primitives   of  the   n   first   degree   differential    equations 


ORDINARY  DIFFERENTIAL  EQUATIONS  219 

obtained  by  equating  to  zero  the  n  factors,  the  product  of  these 
n  primitives 

/i  («,  y,  Ci)/2  («,  2/>  c2)/3  («j  .V,  c3)  •  •  ./n  (a,  y,  cn)  =  0 

includes  all  the  partial  solutions  of  the  given  equation.  Since 
in  each  partial  solution  the  constant  may  have  all  values  from 
+  co  through  0  to  —  oo,  all  possible  values  of  the  partial  solu- 
tions are  included  in  the  product 

/i  fc  y>  <>i)fi  0>  y,  c)  — /„  (x,  y,  c)  =  o, 

where  c  is  an  arbitrary  constant.  This  last  product  is  the 
general  solution  of  the  given  equation. 

Example  I.  —  Solve  St  —  ax  =  0. 
dx2 

Write  the   equation  ^  +  ayY^-a%^  =  0,  and  sol 

\ClX  J  \CIX  J 

the  equations   ^  +  aV  =  0,  ^-#^  =  0.     The  product  of 
dx  dx 

these  solutions,  (y  +  f  a?x%  -f  c)  (?/  —  -§  e**»*  +  c)  =  0,   reducing 

to  (y  +  c)2  —  f  ax3  =  0,  is   the  general   solution  of  the   given 

equation. 

Example  II.  —  Solve  y(^o  +  2x^-- y  =  0. 
dx2  dx 


vc 


Solving  for  %  *— g±2^Z  whence  £**±^Utfe. 
dx    dx         y  y  ±^/x2+y2 

Integrating,   ±  (ar2  -|-  ?/2)^  =  a?  +  c     Hence  the  general  solution 

is   j (as  +  c)  +  (x*  +  2/2)*J  |  (*  +  c)  -  (x2  +  y2)%\  =  0,  reducing  to 

y2  =  2  ex  -f  c2. 

Case  II.  —  Suppose  the  equation  (1)  f(x,  y,  p)  =  0,  where  _p 

stands   for  -%-.  can   be    put    into    the   form    (2)   y  =  F(x,p). 
dx 

Forming  the  ^-derivative  of  (2)  gives  an  equation  of  the  form 


220       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

(3)  p  =  Fx  fx,  p, ' 2 j.    If  the  primitive  of  (3)  is  (4)^  (x,  p,  c)  =  0, 

the  elimination  of  p  from  (1)  and  (4)  is  the  solution  of  (1). 
In  like  manner  if  (1)  f(x,  y,  p)  =  0  can  be  put  into  the  form 

(2)  x=F(y,p),  the  ^/-derivative  of  (2)  is  (3)  1L  =  F(yi^d£ 

P  \         dy 

If  the  primitive  of  (3)  is  (4)  fx  (y,  p,  c)  =  0,  the  elimination  of 
p  from  (1)  and  (4)  is  the  primitive  of  (1). 

Example.  —  Solve  y  =  p2  -f  2ps. 

The   ^-derivative   of  this   equation  is  p  =  2p-^  +  6p2-^, 

dx  dx 

whence    c  -f  x  =  2p  -f-  Sp2.      Hence    x  =  2p-\-3p2  —  c    and 

y  =  p2-\-2ps  for  every  value  of  p  determine  a  pair  of  corre- 
sponding values  of  function  and  variable  of  the  solution  of 
the  given  equation. 

Case  III.  —  If  the  equation  F(x,  p)=0  cannot  be  readily 
solved  for  x  or  p,  try  the  substitution  p  =  xz. 

Example.  —  Solve  x*  + 1£  -  ax^L  =  0. 
dx3  dx 

The   substitution  p  =  xz  gives  x3  -f  x^z2  —  ao?z  =  0,  whence 

az         XT  dy     dy  dz 

*=T+*    ^wp==dx=IzdTx  =  ^ 


dz     l  +  ^dz^l  +  zy        (1  +  z3) 


3     > 


whence  &  =  JgL  A  f_ggL_ \m  ^  -  ^ 

"  dz\l  +  zy 

2z2  —  1 
and  by  integration  y  =  \  a2———  +  \  a2 — - 5  +  C. 

Function  y  and  variable  x  are  now  expressed  in  terms  of  the 
same  quantity  z. 

Case  IV.  —  If  the  equation  F(x,  y,  p)=0  is  homogeneous 
in  x  and  y,  substitute  y  =  xz. 


ORDINARY  DIFFERENTIAL   EQUATIONS  221 

Example.  —  Solve  (2p  +  l)x%y  =  x*p2  +  2  i/i 

The  substitution  y  =  zx,  &  =  z  +  x—  gives  —  ±  -^—  =  0, 
da;  dx  x      z  —  z^ 

whence  ex  -f  (z-  —  l)2  =  0,  and  ex  +  (z*  —  1)~2  =  0  or 
ex2  +  (y*  -  a;*)2  =  0  and  c  +  (y*  -  a;*)"2  =  0. 
Case  V.  —  Clairault's  equation,  y  =  px  -\-f(p)> 
The  x-derivative  of  this  equation  is  p=p  +  #— +/'(.P)— » 

_  CLX  (XX 

dp 
which  reduces  to  (1)  y \x  +f\p)\  =  0.     Equation  (1)  is  satis- 
fied by  (2)  3£  -  o  or  (3)  x  +/'(»=  0.     From  (2)  p  =  c  and 
dx 

the  general  primitive  of  Clairault's  equation  is  y  =  ex  +/(c). 
The  elimination  of  p  from  the  given  equation  and  (3)  gives  a 
singular  solution  of  Clairault's  equation. 

Example  I.  —  Solve  y=px  -\-p  —p3. 

The    general    primitive,    found   by   substituting   p  =  c,   is 
y  =  ex  +  c  —  c3. 

Example  II.  —  Solve  ar*(i/  —  pa;)  =  #p2. 

Multiply  the  given  equation  by  y  and   substitute  u  =  y2, 

^  =  2  2/^.  There  results  wa;2  -  i  a?—  =  i^-  Substituting 
aa?  da;  da;        dar 

x*  =  v,  whence  ^i  =  2x^,  u  =  v—  +  —n,  a  Clairault's  equa- 
da;  dv  dv      dv2 

tion  whose  primitive  is  u  =  cv  +  c2.  Hence  the  primitive  of 
the  given  equation  is  y2  —  ex2  +  c2. 

PROBLEMS 

Solve, 

1.  ^-7^+12=0.  3.   f^xXl+p2). 
dx2       dx 

2.  4  y  =  x2  +  p2.  4.   y2  +  a;#p  —  a^>2  =  0. 


222       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

5.  y  =  xp  +  sin-1  p.  7.    y2(l  —  p2)=6. 

6.  a?  +     ,  P        =  a.  8.    *  =  «*  +  -* 

Vl+i?2  J> 

9.    Sp2y2  —  2  xyp  +  4  ?/2  —  sc2  =  0. 

10.  dy  +  2xy  =  x2  +  y\  14.    arp2  =  l  +  »2. 
eZcc 

11.  a/(p2  +  2)=2p*/3  +  orJ.    15.    x%=l-x. 

dxr 

12.  a;2  +  2/=p2.  16.    x2p2  +  3xyp +  2y2=Q. 

13.  £?£_«  =  0.  17.    ^!(^  +  l)3=l. 
dxr     x  dx2 

18.   ^  +  24-,  =  0. 
dxr  dx 

19.  (ar2-l)^-2^^=l-?/2.     Show  that  the   singular 

dxr  dx 

solution  is  x2  -f  y2  =  1. 

20.  Find  a  curve  such  that  the  area  bounded  by  the  tan- 
gent and  the  coordinate  axes  is  always  a2. 


Art.  77.  —  Ordinary  Equations  in  Three  Variables 

If  the  differential  equation  (1)  Pdx+  Qdy  -J-  Bdz  =  Q  can 
be  solved,  it  may  be  rendered  exact  by  some  factor  fi.     If 

(2)  «=/(»,  y,«)=0  is  the  solution  of  (1),  ^dx+  —  dy-\-—dz=0 

dx  dy  dz 

and  fxPdx  -f  fiQdy  +  fiRdz  =  0  are  identical, 

whence  |^P,  f^MQ,  |«=„fl. 

dx  dy  dz 

The  identities  J*L  =  J*      pL^llL,     &L  ^  J^L  lead 
dxdy      dydx     dxdz      dzdx     dydz      dzdy 

to  the  identities 


ORDINARY  DIFFERENTIAL   EQUATIONS  223 


(5)  "(-s- 


dz  j  dz  dx' 


Multiply  (3)  by  R,  (4)  by  P,  (5)  by  Q,  and  add  the  products. 
There  results 

« *(S;-fMf -SMf -S)-». 

the  condition  under  which  (1)  can  be  solved. 

Example  I.  —  Solve  (y  +  z)dx  +  dy  -f-  dz  —  0. 
Here   P  =  2/  +  z,    Q  =  1,   i2  =  1, 

5P=1   ap=1   5Q=0  i2=0  M=o   *?=0, 

dy  5^        '5a;  dz  dx         '    % 

and  condition  (6)  is  satisfied. 

Considering  x  constant,  the  given  equation  becomes 
dy  4-  dz  =  0,  whence  y  +  z  -\-  X  =  0,  where  X  must  be  so 
determined   that  the    ^-derivative    of    y  -f-  z  -f  X   is    y  -f  z. 

Hence  ^=?/  +  z  =  -X,   i^=-l,  logX  =  c-x,  X=ec~*. 
dx  X  dx 

Finally,  y  +  z  +  ec'x  =  0,  the  solution  required. 
Example  II.  —  Solve 

2(y  +  z)dx  +  (x  +  3y  +  2z)dy  +  (x  +  y)dz  =  0. 
Here  P=2(y  +  z),    Q  =  x  +  3y  +  2z,   R  =  x  +  y, 

^?=2    —  =  2    ^2=1    ^2=1    —  =  1    ^?=1 

dy  dz  dx  dz         '    dsc  dy 

and  condition  (6)  is  satisfied. 


224       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Considering    y    constant,     the     given     equation    becomes 

2  dx         dz 
2(y+z)dx+(x+ y)  dz  =  0,    or  -\ — — -  =  0.     Integrating, 

log  (x  +  y)2  +  log  (y  +  z)  =  log  Y,  whence  (x  +  yf  (y  +  z)=  T, 
where    Y  must  be   so   determined  that   the   ^-derivative   of 
(x  +  y)2(y  +  z)-Y  is   (x  +  3y  +  2z)(x  +  y).     Hence 
2  (?<  +  y)(y+z)  +  (x+yY-^=  (*+3jH-2*)(a>+  y),  ^1=0, 
F=  G.     Finally,  the  required  solution  is  (x  +  y)2(y  -\-z)  =  C. 

PROBLEMS 
Solve,    1.    (y-\-a)2dx  +  zdy  —  (y  +  a)dz  =  0. 

2.  daj  +  cfo/  +  O  +  2/  +  z  +  l)dz  =  0. 

3.  2/z <i#  -f  zx dy  +  a^ cte  =  0. 

4.  (?/  +  z)da;  +  (z  +  a)efa/  +  (a  +  #)dz  =  0. 

5.  zydx  —  zxdy  +  y2dz  =  0. 


CHAPTER  XIII 

OEDINAKY  DIFFERENTIAL  EQUATIONS  OP  HIGHER  ORDER 

Art.  78. — Equations  of  Higher  Order  and  First 
Degree 

Standard  I.  —  The  primitive  of  an  equation  of  the  form 

dnv 

—?-  =f(x)  is  found  by  n  successive  integrations ;  the  primitive 

d^v 
of  an  equation  of  the  form  —%  =f(y)  is  found  by  multiplying 

dv  dx 

both  sides  by  -£  and  integrating,  then  solving  for  —  and 
dx  dy 

integrating  again. 

Example.  —  Solve  J  =  w. 
dxr 

d*y 

Multiply  by        %t,^  =  V% 
Integrating,      |^Y  =  \tf  +  Clf  whence  ^  =  Vs/2  +  2GV 
Solving    or  dx    dx  1 


^y  dy    vy + 2  c\ 

Integrating,         x  =  log  \y  +  vV  +  2Cij  +  <72. 

Standard  II.  —  Equations  of  the  form  f*  +/i  (x)  ^  =/2  (x) 

dx2  dx 

become  linear  by  the  substitution  ~^  =  »,  — \=z-£.. 

dx  dx1     dx 

Q  225 


226       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Example.  —Solve  (1  -  x2) ^ ■  -  x^-  =  0. 

dx2       dx 

Substituting  —=p,  ^  =  ^,  the  given  equation  becomes 
dx  dx2     dx 

l^  =  _a_  Integrating,  ^=C1(1-^2H  Integrating 
p  dx     1  —  x2  dx 

again,  y  =  CX  sin-1  a;  -+-  C2. 

Standard  III.  —  The  equation 

P  dny      p  dn~]y      p  dn~2y  p        Q 

linear  in  y  and  its  derivatives,  the  coefficients  P0,  Plf  P2,-"Q 
being  independent  of  y,  is  called  the  linear  equation  of  order  n. 
Suppose  the  coefficients  P0,  Pl5  P2,  •  •  •  Pn  constant  and  Q  =  0. 

When  ?i = 1,  the  equation  becom  es  P0 -^  -f  P:  2/ = 0  and  ?/ = e     p°x, 

dx 

which  has  the  general  form  y  =  emx.     Substituting  y  —  emx  in 

(1)     p^  +  p^V  +  ...  pny  =  0 
y  J       °dx»        'dx*-1  T 

there  results    (2)     P0  mn  +  I\  mn~l  +  P2  mn~2  +  •  •  •  P„  =  0, 
which  shows  that  ?/  =  ewx  is  a  solution  of  (1)  for  the  n  values 
of  m  which  are  roots  of  (2).     Representing  these  n  roots  by 

mJf  m2,  m3,  ra4,  •••  ran,  y,  =  emix,  ?/2  =  e™2*,  y3  =  em**,  ... ^  _  e«»x 

are  solutions  of  (1).     Consequently 

(3)    y=Cr  emx  +  C2  •  em2*  -f  C3  •  em3X  -\ Cn  •  emnx, 

where  C1}  C2,  C-i,-"Cn  are  arbitrary  constants,  is  a  solution 
of  (1).  Since  (3)  contains  n  arbitrary  constants,  it  is  the  gen- 
eral solution  of  (1).  The  values  of  these  n  constants  become 
known  if  the  values  of  y  and  its  first  n  —  1  derivatives  are 
known  for  some  value  of  x. 


DIFFERENTIAL  EQUATIONS   OF  HIGHER   ORDER      227 

If  the  roots  of  equation  (2)  are  real  and  unequal,  (3)  is  a 
satisfactory  form  of  the  solution  of  (1). 

If  equation  (2)  has  pairs  of  conjugate  imaginary  roots, 
mx  =  a  +  b  V—  Ij  m2  =  a  —  b  V—  1,  or  mx  =  a  +  ib,  m2  =  a  —  ib, 
the  corresponding  terms  of  (3)  are 

=  eax  \  C1  (cos  6a;  +  t  sin  &a?)  +  C2  (cos  6a;  —  t  sin  bx)  \ 

by  Problem  9,  Art.  68.     This  result  may  be  written 

e"*  J  (d  +  C2)  cos  6a;  +  i  (Ci  -  C2)  sin  6a; } , 

which,  by  placing  Cl  =  ^(A  —  iB),  (7,  =  }  (A  +  iB),  becomes 
eax(A  cos  bx+B  sin  6a;).  Finally,  placing  A= Ksin  &,  B=Kcos  k, 
the  result  becomes  K eax  sin  (A;  -f-  6a;),  where  K  and  A;  are 
arbitrary  constants. 

If  equation  (2)  has  equal  roots  ml  =  m2,  (3)  contains  less 
than  n  arbitrary  constants  and  is  no  longer  the  general  primi- 
tive. In  this  case  write  ra2  =  mx  -f  h  and  the  problem  reduces 
itself  to  determining  the  form  of  the  general  primitive 

y  =  C\ .  emi*  -f  C2  •  e(mi+h)x  -f  C3  •  em*x  -\ Cn  •  "*, 

when  h  approaches  zero.     Now 

Cx .  emix-f  C2  •  e(mi+h)x  =  emiX((71  +  C2  •  e»x) 

ss^}Cr1+0/(l-h^+^+-)i 

by  Maclaurin's  series.  When  h  approaches  zero,  this  becomes 
eP*\(Gi  +  G2)  +  C2hx\  =  emiX(A  -f  Bx),  such  values  being  as- 
signed to  the  arbitrary  constants  C±  and  C2  that  Ci  -f  C2  =  ^4, 
C2/i  =  -B  when  7i  approaches  zero. 

If  equation  (2)  has  three  equal  roots,  mx  =  ra2  =  m3,  a  like 
analysis  shows  that  the  corresponding  part  of  the  general 
primitive  is  emiX(A  -f-  Bx  +  Car2). 


228       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

If  equation  (2)  has  equal  imaginary  roots,  for  example 
mx  =  m2  =  (a  -f  ib)  and  ms  =  mi  =  a  —  ib,  the  corresponding 
terms  of  the  general  primitive  are 

eax\  (A  +  Bx)  cos  bx  +  (C  +  Dx)  sin  bx ] . 

Equation  (2)  is  called  the  auxiliary  equation  of  the  differ- 
ential equation  (1). 

Example  I.  —  Solve  ^-  +  6^  +  13  y  =  0. 
dx2        dx 

Substituting  y  =  emx,  the  auxiliary  equation  is  found  to  be 
m2  -f  6 m  -f- 13  =  0,  whence  m1  =  —  3  -f-  2 i,  m2  =  —  S  —  2i  and 
y  =  e'^lA  cos  2 x  +  B  s'm2 x\. 

Example  II.  —  Solve  ^  -  3^  +  4  ?/  =  0. 
dx2        dx2        u 

The  roots  of  the  auxiliary  equation  m3-3m2  +  4  =  0  are 
—  1,  2,  2.     Hence  the  general  primitive  is 

y  =  C  •  e~x+(A+Bx)e2x. 

Standard  IV.  —  Linear  equations  with  second  member  not 
zero. 

Form  the  successive  ^-derivatives  of 

until  either  the  second  derivative  becomes  zero  or  the  elimina- 
tion of  X  from  the  given  equation  and  the  derivative  becomes 
possible. 

Example.  —  Solve  — %  -f  a2y  =  sin  bx. 
dx2        u 

The  second  ^derivative  of  this  equation  is 
dx4  dx2 


DIFFERENTIAL  EQUATIONS   OF  HIGHER   ORDER      229 

Eliminating  sin  bx  from  this  derivative  and  the  given  equa- 
tion  there  results  — ^  +  (a2  +  b2)  — *  -f  «2&1/  =  0.     The  auxiliary 

OtiC  Ota? 

equation  of  this  linear  differential  equation  is 

m*  H-  (a2  +  b2)m2  +  a262  =  0, 

whence  m1  =  ai,  m2  =  —  gw,  m3  =  hi,  m4  =  —  6i.  The  general 
primitive  is  y=  Cx  cos  (ax)  +  C2  sin  (ax)  +  03  cos  (&•#)  -J-  (74  sin  (£■&'). 
This  value  of  y  is  a  solution  of  the  given  second  order  differ- 
ential equation  if  Csa2b  —  C±a2b2  =  0,  —  Csb2  —  CJ)  =  1,  whence 

C3  =  -r_rr2'  C*  =  -j,m    ,  »V      The    re(luired    solution    is, 
1  -f-  0  0(1  +  ^) 

therefore, 
y=  d cos  (ooj)  +  C2sin  (ax) -  — —cos  (6a>) -  ^  sin  (6a?). 

Standard  V.  —  Change  of  the  independent  variable. 

In  the  equation  —^-\-P—+Qy  =  0,  where  P  and   Q  are 
dx2         dx 

functions  of  x,  change  the  independent  variable  from  x  to  z 

by  the  relations  ^  =  ^  .  — ,    ^  =• d^-  f — Y  +  ^ .  *L.    There 
da;     cfo    da;'    dx2      iz2  \dxj       dz    dx2 

results,  (1)  g(4Y  +  ftf*  +  J»»  +  Qy  =  0.    Now  deter- 
dz2\dxj      dzydx2         dx) 

mine  g  so  that   —0+  P  — =0,   whence  (2)  2=  CeSpdx-dx. 
dx2         dx  J 

The  elimination  of  a;  from  (1)  and  (2)  gives  an  equation  whose 
solution  leads  to  the  solution  of  the  given  equation. 

Example.  -  Solve  %  +    *•    &  +  —L  =  0. 

dx2     l  +  x*dx      (1  +  x2)2 

e-fp*x  .  fa  —   I  e  J  l+x2  .  $#  =  tan-1  a;,     whence 

a;  =  tan  2;.  The  transformed  equation  is  — *•  -f  y  =  0,  whence 
?/  =  Ci  •  cos  2  -f-  C2  •  sin  &     Substituting  for  z, 

y  =  d  •  cos  2  +  C2  •  sin  2  =  d  — =  +  C2 


V 1  +  x2  Vl  +  x2 


230       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


PROBLEMS 

Solve, 

1.  g  =  sin»x.  12.   &=%■  +  y. 
aar  dx-     dx 

2.  <§M  +  5<k  +  iy  =  0.  13.   f|_3^  +  2/  =  0. 
aar        dx  dx3        dx 

3.  rfjgUl.  14.    fiU^.,  +  1. 

da4  da;2     * 

4.  f^  =  0.  15.    &  +  ,-«»» 

da;2      y2  dx2 

5.  ^  +  6^  +  9j,  =  0.  16.    ?l-3'*l  +  2y  =  x.e' 
dor        dx  dxr        dx 

6.  ^L]l-m2y  =  0.  17.    ^  +  3 2/  =  sin  (ma;). 
dx2         a  dx2        u  s     ; 

7.  — ?  =  x2  •  sin  x.  18.   —^  +  4?/  =  cos  (we). 
clar  dx2 

dx3     dx2        dx  dx2        dx 


9.  ^+2^_82/=o.      20.  ■a_*-+4*-* 

aar        dx*  dx*     dx 

io.  g^8fij+4f-ft      2i.  gw§*v+i. 

dx*        aar  ■        dx'      \dxj 

11.    &-'•'. V.  22.    **?  +  &=*. 

dar  dar     da; 

Art.  79.  —  Symbolic  Integration 

If  u,  v,  w  are  any  symbols  whatever  obeying  the   funda- 
mental laws  of  ordinary  algebra,  namely : 


DIFFERENTIAL  EQUATIONS   OF  1IIG1IER   ORDER     231 

the  associative  law,  u  +  v  4-  w  =  (u+v)+w,  uvw  =  (uv)w  ; 
the  commutative  law,  u-\-v  +  w  =  v  +  u  +  w,  uvw  =  vuw ; 
the  distributive  law,  (u  -f-  v)  w  =  uw  -\-  vw  ; 
the  law  of  indices,  umun  =  um+n ; 

any  algebraic  transformation  of  expressions  involving  u,  v,  w 
gives  a  valid  result. 

Denoting  the  operation  of  forming  the  first  derivative  by  D, 
that  of  forming  the  second  derivative  by  D2,  that  of  forming 

the  third  derivative  by  D3,  •••,  so  that  Dy  =  %    D2y  =  ^, 

d3v 
D*y  =  — ^,  •••;    and  denning  the  symbol  D~l  by  the  equation 

D(D~Jy)  =  y,   whence   D1   is   equivalent  to   the   symbol   of 

integration  I  ,  it  follows  that : 

I.  D  •  Dy  =  D2y,       D  •  D2y  =  &y,       D"1  •  D2y  =  By, 

D~2  •  Dy  =  D~ly ;   that  is,  the  law  of  indices  of  algebra  holds 
for  D  affected  by  integral  exponents. 

II.  (a  +  bD-  cD2) y=\(a  +  bD)-cD2\y;  that  is,  the 
associative  law  of  algebra  holds  for  D  combined  with 
constants. 

III.  (aD  +  bD2) y  =  D(a  +  bD) y ;  that  is,  the  distributive 
law  of  algebra  holds  for  D  combined  with  constants. 

IV.  (a  +  bD)  y  =  (bD  +  a)  y,  aDy=D  ay ;  that  is,  the  com- 
mutative law  of  algebra  holds  for  D  combined  with  constants. 

Since  the  operator  D  obeys  the  fundamental  laws  of  algebra, 
the  result  of  any  algebraic  transformation  of  expressions  con- 
taining D  and  constants,  is  valid.  For  example,  by  Maclaurin's 
series, 

Hence  f(D)  y=%An>  D»y.     If  /(D)  y  =  X,  y  = -^  X    This 

is  the  definition  of  the  operator  inverse  to  f(D). 

f(D) 


Hence  — ± (2  x3  -  23  x2  +  49  x  -  31)  =  2  a?  -  5  x2  +  7*. 

2>2-3i>  +  lv  7 


232       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

For  instance, 

(Z)2  -  3  D  +  1)  (2  ar>  -  5  x2  +  7  x)  =  2  x3  -  23  x2  +  49  a?  -  31. 
1_ 
*  + 

Example  L  — Show  that  (D2  -  D-2)y  =  (D  +  l)(D-2)y. 
(^_i,_2)!,sg_|_2y.  (D_2)ys|_2s,  and 

(i,+i)(i>_2)ys(i,+i)(|-2y)sg-2|+|_2!, 
_«S_^_2«. 

do2     da; 

Example  II.  —  Show  that  /(D)  eax  =/(a)  eaz. 

/(D)  =  2 A#"  and  ADMeaac  =  Ananeax. 

Hence    f(D)eax=f(d)eax.     Inversely  — =—  eax  =  -=—eax. 
K    }  W  J  /(D)         /(«) 

Example  III.  —  Show  that  — — =- — -  ?/  =  [  — i — 5— ]  y. 

D2-D-2J     \D+1     D-2Jy 

Performing  the  operation  D2  —  D  —  2  on  both   sides   of  the 
assumed  identity, 

(j)2-D-  2) 1 y  =  (D2-D-  2)( — i i— \ 

or  2/=  {  —  -|-D-f-f-f-iD-}-£|?/==2/,  which  proves  the  assumed 
identity  correct. 

Example  IV.  —  (D2-2D  +  2)y  =  6x-9x2  +  2xs.    Find?/. 

y  = - (6x-9x2  +  2xi) 

9      D2-3D  +  2V  T     J 

=  -J—(6x-9x2  +  2x3)-—l—(6x-9x2  +  2xi) 
1  —  1)  £  —  D 

=  (1  +  D  +  D2  +  D3  +  D4  +  •••)  (6x-  9X2  +  .2  a?) 
(\  ,  D  ,  D2  .  D3  .  D4  .       N,*         Q   2  ,  0   ,, 

=  ar>. 


DIFFERENTIAL  EQUATIONS   OF  HIGHER   ORDER     233 

This  is  a  particular  value  of  y  since  it  does  not  contain  arbi- 
trary constants. 

Example  V.—  y  =  [0].     Find  y. 

This  is  equivalent  to  (D  —  2)  y  =  0,  which  is  the  same  as 
the  linear  equation  -^  —  2  y  =  0.     Whence  y=e2x. 

CLX 


PROBLEMS 

1.  Show  that  /(D)  [ea*X]  =  eaz  -f(D  +  a)  [X],  and  conse- 
quently — —  [eaxX]  =  eax [XI. 

4  J  /(I>)    L         j  /(Z>  +  a)L     J 

2.  (Z>2  +  D  +  l)2/  =  ea;^.    Showthat2/  =  ex^-»2  +  |a;  +  ^\ 

3.  Show  that  f(D2)  sin  (ma?)  =/(—  m2)  sin  (ma;)    and  conse- 
quently — - — -  sin  (ma;)  =  — sin  (mx). 

4.  Show  that  /(D2)  cos  (mx)  =  /(— m2)  cos  (ma;)   and  conse- 
quently   —  cos  (mx)  =  — - —  cos  (mx). 

4         y  f(D2)       V      ;     /(-m2)        V      ; 

5.  Show  that  ^^«  =  -£-* 

6 .  Show  that   — e2x  sin  x  =  —  e2x  sin  x. 

D2-4Z)  +  4 

8 .  Show  that  — — -  ex  sin  x  =  e*  (sin  a;  —  cos  a;). 

9.  (D  -  m)  ?/  =  0.     Show  that  y  =  ?-i —  [0]  =  e~. 

U  Tib 

10.   Show  that  y  =       °       [0]  =  emx  ((7,  -f  (%*). 

(  jl/  —  //I ) 


234       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 


Art.  80. — Symbolic  Solution  of  Linear  Equations* 

Writing   the   linear   equation    P0^  +  pi^ri+  —  J\V  =  0 

ctx  ax 

with  constant  coefficients  and  second  member  zero  in  the  form 
(P0D»  +  P,D^  +  P2D"-2  +  -  Pn)  V  =  0, 

V  =  P0D»  +  Pj*=*  +  PnD»~2  +  ...  Pn[°]* 

Factoring,  P0Z)n  +  P^D""1  +  P2Dn~2  + —Pn 

=  (D-  mx)  (Z)  -  m2)  (Z>  -  m3)  ...  (D  -  mn). 

Decomposing  into  partial  fractions, 

y=7i^-[0]+7^-io]+7r^-[0]  +  ...+-S_[d]) 

D—rrii           D—m2            D—mz  D—mn 

hence,        ?/  =  CieTOl*  -f  Cge™**  -f-  C3eTO3X  H h  Cnemnx. 

C 
If  mj  =  m2,  a  partial  fraction  of  the  form  occurs. 

The  value  of  - — - JO]  is  (Cj  +  C^c)e^x. 

(D  —  ?%)* 

In  general,  if  m1  =  m2  =  m3=  •  •  •  rar,  a  fraction  of  the  form 

(7  C 

occurs.     The  value  of  — [0]  is 


(D  -  m1)r  (D  -  mj 

(Ci  +  C^  +  C3«2  +  -  CU**"*  +  Cr_^-2  +  0,0  *•*". 

If  mx  =  m2  =  a  -j-  bi,  m3  =  ra4  =  a  —  bi,  the    corresponding 
terms  of  the  solution  of  the  differential  equation  are 

reducing  to  {Axx  +  B{)  cos  (bx)  -+-  (A2  +  B&)  sin  (&#). 

The  solution  of  (1)  P0^l1 -{.  p1^—]i  +  ...  +P«=X,  with 

*  Maclaurin  introduced  the  symbolic  solution  of  differential  equations. 


DIFFERENTIAL  EQUATIONS   OF  HIGHER   ORDER      235 

coefficients  constant  and  right-hand  member  a  function  of  x, 
is  y  =  Y+  u,  where   Y,  called  the  complementary   solution 

of  (1),  is  the  solution  of  P0^  +  P^-^  +  •••  +  Pjl  -0,  and 

dxn  dxn~x 

u  is  any  particular  solution  of  (1).     For  substituting  y  =  Y  +  u 
in(l)!(i>0|^+P1|J+...+P„r 

which  is  a  true  equation  by  the  hypothesis. 

d4v 
Example  I.  —  Solve  —?-—  y  =  x4. 
dx4     9 

Writing  the  complementary  equation  (D4  —  1)  Y  =  0,  or 
(D  +  1)(D-1)(D+V^1)(D-a/~^1)Y=0,  the  complemen- 
tary solution  is  found  to  be  y=01ex+C2e~x+C3sin;c4-(74Cossc. 
The  particular  solution  is 

u  =  —^—x4  as  (-  1  -  D4  -  B* )  x4  =  -  x4  -  24. 

D4  —  1  J 

Hence  the  general  solution  of  the  given  equation  is 

y  =  dex  -f  (72e_a:  -f  (73 sin x -\-  C^cosx  — x4  —  24. 

Example  II.  —  Solve  ^-2^  +  2/  =  ^. 
cZar        dec 

The  solution  of  the  complementary  equation 

(D2-2D  +  1)Y=0  is    F=  6»(Ci+Ci»). 
The  particular  solution  is 
1 


D2-2Z>  +  1 


(a*?3*) 


gji  = 2 = ^EEC3*- 


(Z>  +  3)2  _  2  (Z>  +  3)  -f  1  (2  +  £>)2 


236        DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

The  general  solution  is 

y  =e-(G1  +  C&)  +  \e**(2x>  -  4  a  +  3). 

Example  III.  —  Solve  ^  +  ^  +  y  =  sin  (2  a). 
dor2     das 

The  solution  of  the  complementary  equation 

(D2  +  D  +  1)Y=0   is    F  =  e^C1cos^a>+(72sin^A 
The  particular  solution  is 

u  =  — sin  (2  a;)  = sin  2  a  =     „       sin  2  a? 

=  _1J5(JD  +  3)sin2ic=  -  ^ (2  cos 2 as  +  3 sin 2  a?). 
Hence  the  general  solution  is 

y  =  e~\(cx cos ^x  +  <72 sin^a)  -  TV (2  cos  2  a;  +  3 sin  2 as), 

PROBLEMS 

Solve, 

1.  ?LM  +  y  =  l+x  +  x>.  e.   g-2|M2/=e*cosz. 

2.  §-2^  +  2/=e*.  7.  g+32/  =  sin^. 
c?ar        da;  dar  T 

3.  3^  +  2/  =  cosa;.  8.   -^+ 4?/  =  cos(naj). 
dar  dar 

4.  ^-3^  +  2y  =  a»ta.  9.   ^2{  +  y  =  xsm2x. 
dx2        dx  dx2 

5.  ^o  +  4?/  =  a;sin2a;.  10.    ^0  +  4 v  =  2 aj3 sin2 x. 
dx2  dx2 

11.   ^  +  2^+2y  =  eaesina>  +  cos». 
dar        da; 


12.   f^4  +  2^+2/  =  a;2cosa;. 


da;4        dx2 


DIFFERENTIAL  EQUATIONS   OF  HIGHER   ORDER     237 

Art.  81.  —  Systems  of  Simultaneous  Differential 
Equations 

Let  Pc-^-=Q,   P1-^-=Q1,    where    x    is    the    independent 
dx  dx 

variable  and  y  and  z  are  the  dependent  variables,  and  P,  Q, 
Plf  Qi  are  functions  of  x,  y,  z.  It  may  be  possible  by  combin- 
ing the  given  equations  with  their  derivatives  to  obtain  an 
equation  in  which  one  dependent  variable  and  its  derivatives 
do  not  appear. 

Example.  — Solve  ^-7x  +  y  =  0,  ^-2x-5y=0. 
dt  dt  9 

Form  the  ^-derivative  of  the  first  equation, 

d2x     „dx     dy  _r. 
di?        dt dt~ 

and  eliminate  y  and  -^  from  this  equation  and  the  two  given 
equations.    There  results  the  linear  equation 

^-12— +  37a  =  0, 

dt2  dt  ' 

whence  x  =  e6'(Ci  •  cos  t  +  C2  •  sin  t).  Substituting  this  value 
of  x  in  the  first  of  the  given  equations, 

y  =  e6<  j(Q  -  Q  cos*  +  (ft  +  C2)  smt\. 

If  the  given  equations  can  be  written  in  the  form 

dx  _  dy  _  dz 
P~  Q~B' 

each  of  these  equal  ratios  equals  ^  +  m(fy  +  ncfe   and  the 

H  H  IP+mQ  +  nR  ' 

last  ratio  implies  the  same  relation  between  x,  y,  and  z  as  the 

three  given  ratios.      If  I,  m,  n  can  be  so  determined  that 


238       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

IP  +  mQ  -f-  nil  =0,  it  follows  that  I dx  +  m  dy  +  n  dz  =  0. 
The  integral  of  the  last  equation  is  an  integral  of  the  given 
system  of  equations. 

Example.  -  Solve  ^  =  ^  =  ^. 
y2z       xz     i/ 

ttt  -i    ®dx     dy     dz     Ixdx  +  mdy  +  ndz 
Write  — —  =  -»=—=  \ *J        . 

Placing  l  =  —  x,  m  =  y'2,  n  =  0,  Z?/22  +  m#2  +  w?/2  =  0,  whence 

—  x2dx-\-y2dy  =  0  and  y3  —  Xs  =  Ci. 

Placing  Z  =  —  1,  m  =  0,  n  =  z,  ly2z  +  ra#z  +  ny2  =  0,  whence 

—  xdx  -{-zdz=0  and  z2  —  x2  —  C2. 


PROBLEMS 

Solve'1-f+l+2a;+2'=0'l+5a;  +  32'=0- 

„     dx  _dy  _  dz 
x2      y2      xy 

32/  +  4z     2y  +  5z 
5*   5-3aj  +  42/  +  3  =  0,g+aJ4-y  +  5  =  0. 

eft2        eft  '  dt2        dt  7    * 


CHAPTER  XIV 

PARTIAL  DIFFERENTIAL  EQUATIONS 

Art.  82.  —  Formation  of  Partial  Differential 
Equations 

Example  I.  —  Form  the  partial  differential  equation  of  the 
system  of  spheres  (1)  (x  —  a)2  -f- (y  —  b)2  -f  z2  =  R2,  whose 
centers  (a,  b,  0)  lie  in  the  XF-plane  and  whose  radius  is 
constant. 

Consider  z  the  dependent,  x  and  y  the  independent  variables, 

and  denote  —  by  p,  —  by  q. 
dx    J       dy    J  H 

Differentiating  (1)  partially  with  respect  to  x, 

(2)  («-a)+zp  =  0; 
differentiating  (1)  partially  with  respect  to  y, 

(3)  (y-b)+zq  =  0. 

Eliminating  a  and  b  from  (1),  (2),  (3),  z2(l  +p2  +  q2)=  R2, 
the  partial  differential  equation  required. 

Example  II.  —  Form  the  partial  differential  equation  of 
the  expression  z  =  fa(y  +  ax)  -f  cf>2(y  —  ax),  where  <f>x  and  <f>2 
are  arbitrary  functions. 

Writing  the  given  expression  z  =  (fr^v^-t- <f>2(v2),  which 
requires    that    vx  =  y  +  ax,    v2  —  y  —  ax,    and    differentiating 

239 


240       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

twice  in  succession  partially  with  respect  to  x  and  also  with 
respect  to  y, 

g  =  a.^00  -  a^(v2),  g  =  a2  •  +»,(«,)  +  a2  •  *«,(«<>> 

d2Z  32Z 

By  division,  — -  =  a2  •  — — ,  the  partial  differential  equation 
dx2  By2 

required. 

Observe  that  partial  differential  equations  result  either 
from  the  elimination  of  arbitrary  constants  from  a  function 
and  its  successive  partial  derivatives,  or  from  the  elimination 
of  arbitrary  functions  from  an  expression  and  its  successive 
derivatives. 

PROBLEMS 

Form  the  partial  differential  equations  of  the  following 
expressions : 

1.  z  =  ax  +  -  +  b.  3.   z  =  ax  +  by -\- ab. 

2.  z  —  ax  +  dly2-\-b.  4.   z  —  <£(?/  -f-  mx). 

5.  y  —bz  =  cf>(x  —  az). 

6.  z  +  ay  +  bx=<j>\(x-a)2+(y-P)2  +  z2\. 


Art.  83. — Partial  Differential  Equations  of  First 

Order 

Standard  I.  —  Equations  of  the  form  F(p,  g)=  0. 

Try  a  solution  of  the  form  z  =  ax  +  by  -f  c.  Since  p  =  a 
and  q  =  b,  z  =  ax-\-by  +  c  is  a  solution  of  F(p,  #)=0  if 
F(a,  b)=0.     Denoting  by  /(a)  the  value  of  b  obtained  from 


PARTIAL  DIFFERENTIAL  EQUATIONS  241 

the   equation   F(a,  6)  =  0,    the    solution   of    F(p,q)  =  0   is 
z  =  ax+f(a)'y  +  c. 

Example.  —  Solve  p2  +  q2  =  m2. 

Here  a2  +  b2  =  m2,  whence  6  =  Vm2  —  a2,  and  the  solution  is 
z  =  ax  -fVm2  —  a2  •  y  +  c. 

Standard  II.  —  Equations  of  the  form  F(z,  p,  q)  =  0. 

Try    a    solution    of     the     form     z  —  <j>  (x  +  a?/).      Writing 

,  /  N  .  dz  dv      dz  -,  dz  dv         dz 

z=4(v),    v  =  x+ay,   p  =  —  ~  =  —    and    ?  =  — .   =«— . 

ay  ax     av  dv  oy        dv 

By  substituting  these  values  of  p  and  q  the  given  equation 

becomes  in  z,  — ,  a  —  )=0,   an  ordinary  differential  equation 
V     dv      dv) 

whose  solution  leads  to  the  solution  of  the  given  partial  differ- 
ential equation. 

Example.  —  Solve  9  (p2z  +  q2)  =  4. 

Writing  z  —  cf> (v),  v  =  x+ay,  whence  p  ==  —  and  q  =  a—, 

dv  dv 

the  given  equation  becomes 

\Zdv2+adv2)-*>  and  dv~3^f^' 
Integrating,  }  (v  +  c)  =  |  (a  +  a2)*  or   (x  +  ay  +  c)2  =  (*  +  a2)3. 

Standard  III.  —  Equations  of  the  form  F1(x,p)  =F2(y,  q). 

Assume  (1)  Fx  (x,  p)  =  a,  (2)  F2  (y,  q)  =  a,  where  a  is  an 
arbitrary  constant.  Integrating  (1),  z  =/x  (x,  a)  +  Y,  where  T 
represents  the  terms  of  z  which  do  not  contain  x\  integrating 
(2),  z  =  f2  (y,  a)  +  X,  where  X  represents  the  terms  of  z  which 
do  not  contain  y.  Hence  z  =/i(#,  a) +/2(2/,  a)  +  C  is  the 
required  solution. 


242       DIFFERENTIAL  AND  INTEGRAL   CALCULUS 

Example.  —  Solve  p2  +  q2  =  x  +  y. 

Write  this  equation  p2  —  x  =  y  —  q2  —  a,  whence 

p  =  |*  =  («  +  »)*  and  g=^.=  (y_a)l. 
ox  oy 

Integrating,     z  =  |(a+  ^)f  +  T,    z  =  ±(y-a)*  +  X, 

and  z  a  |  (a  +  a)1  +  |  (2/  -  a)?  +  Q 

the  required  solution. 

Standard    IV.  —  The    analogue    of    Clairault's    equation 
z=px  +  qy  +  <f>(p,  Q)-      The    solution    of    this    equation    is 

z  =  ax  +  by  +  <f>(a,b),  for  ~=p  =  a,  -^-  =  q=b. 

ox  oy 

Example.  —  Solve  z  —px  +  qy  +  (1  -j-p2  +  q2)*. 
The  solution  is  z  =  asc  +  by  +  (1  +■  a2  4-  62)^. 

Standard  V.  —  Lagrange's  solution  of  (1)  P— +  Q  —  =j?, 

bx  by 

where  P,  Q,  R  are  functions  of  x,  y,  z. 

Suppose    (2)   u  =  F(x,  y,z)  =  a    to    be    a    solution   of  (1). 

Differentiating  (2)  partially  with  respect  to  x  and  yt 

•o\    bu      bu  bz_  _  r.       ,.,    bu      bubz  _  ^ 
d#      5g  bx  by      bz  by 

Solving  (3)  and  (4)   for  -^  and  -^  and  substituting  in  (1), 

ox  oy 

there  results  (5)  P~  +  Q^  +  u£H  =  0.     Hence  a  solution  of 
bx  by  bz 

(1)  is  also  a  solution  of  (5),  and  conversely. 
Writing  the  system  of  equations 

bu  ,     .  bu  ,     .  bu  , 

— -  ax  -f  — -  <%  H dz 

dx_dy_dz      bx  by  bz 

P~  Q~~  R~  p^iq^,^^' 
bx  Cy  bz 


PARTIAL  DIFFERENTIAL  EQUATIONS  243 

it  is   evident  that  if  u  =  a  is  a  solution   of  the  system  of 

fl'7*  fill  fl% 

equations     —  =  -^  =  -— -,    it    must    also    be    a    solution    of 
Jr        ty      It 

pdu+  qdJijrRdjL  =  oi  and  consequently  of  P^+Q?1=B. 
ox  By  Bz  Bx  By 

If  u  —  a  and  v  =  b  are  two  independent   solutions  of  the 

fJnf         rffi         nSL 

system  of  equations  —  =  -^  =  — ,  f(u,  v)  =  0,  where  /  repre- 

JT  (q£  a 

sents  an  arbitrary  function,  is  also  a  solution  of  (1).     Fox 
P£f(«,  v)  +  Q±f(u,  v)  +  B !/(«,  v) 
jy  B  j,,       v    Bit  t    n  9  »t       v    dv  ,    *  d  .,       ,    Bu 

s  PYuf{u'  v)'Tx  +  Pd-/(w>  V)'TX+  QYuf{u' v)  ■  8i, 

+  Q  Tvf{U'  V)-6y+  STuf(U'  ?>  *  5  *  BlTvf(u>  ^  *  & 

+if^{pt+  QTy+Rdi}=° h*  h™othesis- 

The  solution  f(u,  v)  =  0  may  be  written  u  =  <f>  (v),  where  <f> 
represents  an  arbitrary  function. 

Bz  Bz 

Example.  —  Solve  xz  •  —  +  yz  •  —  =  xy. 
ox  By 

The  system  of  equations  —  =  —  =  —   is  satisfied  by  -  =  a 
•  *  xz      yz      xy  y 

and  xy  —  z^—b.     Hence  the   general  solution  of  the   given 
equation  is  xy  —  z2=  <f> 


PROBLEMS 

Solve, 

1.  pq  =  k.  2     q  =  xp  +  p'\  3.    z  =  px  -f  qy  +  pq. 


244       BIFFEBENTIAL  AND  INTEGRAL   CALCULUS 


I        I      r»  n        dz    .      dz 

4.  p*  +  q*  =  2x.  9.    z  —  +  V-Q-  =  x. 

5.  p>  +  q>  =  nPq.  io>  B*+f*  +     a 


6.  p2  =  z2(l—pq). 


dx         dy 


U.    ^^_^^  +  2/2  =  0> 

7.  i?(l  +  g)  =  g».  5aj         % 

8.  0=^  +  ^  +  3pV.  12'    ^Yx+y2Y  =  Zi' 

Art.  84.  —  Linear  Equations  of  Higher  Order 
Standard  I.  —  All  derivatives  of  the  same  order. 

Example.  —  Solve  ^9  +  3-^-  +  2p-  =  0. 
ox2        ox  dy        dy2. 

Assume   z  =  $(y  +  ma),  where   <f>   represents   an   arbitrary 

dz 
function.      Writing    z—  <f>(v)   and    v  =  y  +  wiaj,  —  =  m<f>'(;u), 

B—^^  S=w*"(w)'  ir*'^  |r  *"<«>•  Sub- 

stituting  in  the  given  equation,  (m2  +  3m  +  2) <£"(V)  =  0. 

Hence  z  =  4,(2/  +  ma)  is  a  solution  of  |^  +  3-^-  +  2—  =  0 

oar        da;  01/        dy2 

if  m2  +  3  m  +  2  =  0,  that  is  if  w  =  —  1,  m  =  —  2,     The  required 
solution  is  therefore  z  =  ^(y  —  x)-{-  cf>2(y  —  2  #). 

Standard  II.  —  General  linear  equation. 

Example. -Solve  *L-**-3*  +  s£ -« 

oar      o?/^         da;        oy 

Assume    2  =  *■"+■»,    whence     —  =  memx+ny,     —  =  m2emx+By, 

da;  '     dx2 

dz  d27 

a?/  dy2 


PARTIAL  DIFFERENTIAL   EQUATIONS  245 

Substituting  in  the  given  equation, 

(m2  —  n2  —  3  m  +  3  7i)emx+nv  =  0. 

Hence  z  =  emx+nv  is  a  solution  of  the  given  differential  equa- 
tion if  m2  —  n2  —  3  m  -f  3  ?i  =  0.  Solving  this  equation  for  m, 
m  —  ny  m  =  3  —  n.  Hence  zx  =  en(*+,°  and  z2  =  e3*  •  en{y~x)  are 
solutions  of  the  given  equation  for  all  values  of  n,  and  in 
general  %  =  2J.neM(*+y)  +  e&2.Bnen(y~x),  where  n,  An1  and  i?„  are 
arbitrary  constants  is  a  solution  of  the  given  equation.  Since 
^Anen{x+y)  and  %Bnen{y~x)  are  arbitrary  functions  of  x  +  y  and 
2/  —  a?  respectively,  this  solution  may  be  written 

PROBLEMS 

Solve, 

1.  f^  +  5_^_+6  —  =  0.  3.    ^=a2^- 

Sic2        da?d?/        3?/2  3£2  3a?2 

32z        32z        3z  _  3z  _  ^  c^z  _    32z        3z        _  ^ 


6a;2     3x3?/     3a;     3#  dx2     dxdy     By 

6.   2^-3-^--2^ 
dx2        dx  dy        By2 


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process,  however  simple,  is  deemed  unworthy  of  clear  explanation.  Where  it  seems 
advantageous,  a  rule  is  given  after  the  explanation.  .  .  .  Mr.  Lock's  admirable 
•Trigonometry '  and  the  present  work  are,  to  our  mind,  models  of  what  mathematical 
school  books  should  be."  —  The  Literary  World. 


FOR   MORE   ADVANCED   CLASSES. 

ARITHMETIC. 

By  CHARLES  SMITH,  M.A., 

Author  of  "  Elementary  Algebra"  "A  Treatise  on  Algebra" 

AND 

CHARLES    L.  HARRINGTON,  M.A., 

Head  Master  of  Dr.  J.  Sach's  School  for  Boys,  New  York. 

1 6mo.    Cloth.    90  cents. 

A  thorough  and  comprehensive  High  School  Arithmetic,  containing  many  good 
examples  and  clear,  well-arranged  explanations. 

There  are  chapters  on  Stocks  and  Bonds,  and  on  Exchange,  which  are  of  more 
than  ordinary  value,  and  there  is  also  a  useful  collection  of  miscellaneous  examples. 


THE   MACMILLAN   COMPANY, 

66  FIFTH  AVENUE,  NEW  YORK. 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

RENEWALS  ONLY— TEL.  NO.  642-3405 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 

Renewed  books  are  subject  to  immediate  recall 


'JUN  1 5  1971 


ftECO  to  Jim  4    71-5PN177 


LD2lA-60m-3,'70 
(N5382sl0)476-A-32 


General  Library 

University  of  California 

Berkeley 


LD21-100m-12,'43  (8796s) 


W306070 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


